What is the slope of the line normal to the tangent line of #f(x) = tanx-secx # at # x= (5pi)/12 #?
Slope of the normal to the tangent line is
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To find the slope of the line normal to the tangent line of ( f(x) = \tan(x) - \sec(x) ) at ( x = \frac{5\pi}{12} ), we first need to find the slope of the tangent line at that point, and then find the negative reciprocal of that slope, which will give us the slope of the normal line.
- Find the derivative of ( f(x) ) using the quotient rule:
[ f'(x) = \frac{d}{dx}(\tan(x)) - \frac{d}{dx}(\sec(x)) ]
[ = \sec^2(x) - \tan(x)\sec(x) ]
- Evaluate ( f'(x) ) at ( x = \frac{5\pi}{12} ):
[ f'\left(\frac{5\pi}{12}\right) = \sec^2\left(\frac{5\pi}{12}\right) - \tan\left(\frac{5\pi}{12}\right)\sec\left(\frac{5\pi}{12}\right) ]
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Calculate the value of ( f'\left(\frac{5\pi}{12}\right) ).
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The slope of the tangent line at ( x = \frac{5\pi}{12} ) is ( f'\left(\frac{5\pi}{12}\right) ).
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The slope of the normal line is the negative reciprocal of the slope of the tangent line.
Therefore, the slope of the line normal to the tangent line of ( f(x) = \tan(x) - \sec(x) ) at ( x = \frac{5\pi}{12} ) is the negative reciprocal of ( f'\left(\frac{5\pi}{12}\right) ).
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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.
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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