# What is the slope of the line normal to the tangent line of #f(x) = secx+sin(2x-(3pi)/8) # at # x= (11pi)/8 #?

The slope of the normal line is given by

The derivative is given by

so the slope of the normal line is given by

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To find the slope of the line normal to the tangent line of ( f(x) = \sec(x) + \sin\left(2x - \frac{3\pi}{8}\right) ) at ( x = \frac{11\pi}{8} ), first, find the derivative of the function. Then, find the derivative evaluated at ( x = \frac{11\pi}{8} ) to get the slope of the tangent line. The slope of the line normal to the tangent line is the negative reciprocal of the slope of the tangent line.

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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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