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How do you find the derivative of #y=ln(sec(x)+tan(x))#?

Answer 1

Ezekiel Alexander

This question has already been answered on this following link :

What is the derivative of #y=ln(sec(x)+tan(x))#?

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Answer 2

Isabella Ackerman

To find the derivative of ( y = \ln(\sec(x) + \tan(x)) ), you can use the chain rule.

The derivative of ( \ln(u) ) with respect to ( x ) is ( \frac{1}{u} \cdot \frac{du}{dx} ).

Let ( u = \sec(x) + \tan(x) ).

First, find ( \frac{du}{dx} ), the derivative of ( u ) with respect to ( x ), using the sum rule and the derivatives of secant and tangent functions.

( \frac{du}{dx} = \sec(x)\tan(x) + \sec^2(x) ).

Then, substitute ( u ) and ( \frac{du}{dx} ) into the derivative formula for ( \ln(u) ).

( \frac{dy}{dx} = \frac{1}{\sec(x) + \tan(x)} \cdot (\sec(x)\tan(x) + \sec^2(x)) ).

Simplify if necessary.

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Answer from HIX Tutor

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.