# What is the slope of the polar curve #f(theta) = theta - sec^3theta+thetasin^3theta # at #theta = (5pi)/8#?

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To find the slope of the polar curve at aTo find the slope of the polar curve ( f(\To find the slope of the polar curve at a given pointTo find the slope of the polar curve ( f(\theta)To find the slope of the polar curve at a given point, (To find the slope of the polar curve ( f(\theta) =To find the slope of the polar curve at a given point, ( \To find the slope of the polar curve ( f(\theta) = \thetaTo find the slope of the polar curve at a given point, ( \thetaTo find the slope of the polar curve ( f(\theta) = \theta -To find the slope of the polar curve at a given point, ( \theta = \To find the slope of the polar curve ( f(\theta) = \theta - \To find the slope of the polar curve at a given point, ( \theta = \fracTo find the slope of the polar curve ( f(\theta) = \theta - \secTo find the slope of the polar curve at a given point, ( \theta = \frac{To find the slope of the polar curve ( f(\theta) = \theta - \sec^To find the slope of the polar curve at a given point, ( \theta = \frac{5To find the slope of the polar curve ( f(\theta) = \theta - \sec^3To find the slope of the polar curve at a given point, ( \theta = \frac{5\To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\thetaTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta)To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) +To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8}To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \thetaTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ),To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), youTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sinTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'llTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need toTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to findTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\thetaTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find (To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta)To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) )To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \fracTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) atTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dyTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \thetaTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{dTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta =To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\thetaTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \fracTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}}To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} \To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} )To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and (To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \fracTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ),To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), firstTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{dTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first,To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\thetaTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate theTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}}To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivativeTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} \To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative ofTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} )To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of (To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) andTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( fTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and thenTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate (To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\thetaTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) \To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \fracTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) )To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) withTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect toTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dxTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to (To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}}To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} \To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \thetaTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) usingTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta \To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using theTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) usingTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formulaTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using theTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula (To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polarTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \fracTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinateTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula forTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\fracTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives.To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. ThenTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dyTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then,To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substituteTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{dTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute (To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\thetaTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \thetaTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta =To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\fracTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dxTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\piTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{dTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\thetaTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}}To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} \To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} \To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} )To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ).To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) intoTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). HereTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into theTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, (To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expressionTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( yTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression toTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y =To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to findTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = fTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find theTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slopeTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\thetaTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slope atTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\theta)To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slope at thatTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\theta) \To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slope at that pointTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\theta) )To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slope at that point.To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\theta) ) andTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slope at that point.To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\theta) ) and (To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slope at that point.To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\theta) ) and ( x = \theta ). So, ( \frac{{dy}}{{d\theta}} ) is the derivative of ( f(\theta) ) with respect to ( \theta ), and ( \frac{{dx}}{{d\theta}} ) is simply 1. After finding these derivatives, plug in ( \theta = \frac{5\pi}{8} ) to get ( \frac{{dy}}{{dx}} ), which gives you the slope of the curve at that point.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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