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What is the derivative of #cosx^tanx#?

Answer 1

First find the derivative of #y=x^{tan(x)}# by taking the natural log of both sides to get #ln(y)=ln(x^{tan(x)})=tan(x)ln(x)# and then differentiating with the Chain Rule (on the left) and Product Rule (on the right to get #\frac{1}{y}\frac{dy}{dx}=sec^{2}(x)ln(x)+(tan(x))/x#. Now multiply both sides by #y=x^{tan(x)}# to get #dy/dx=x^{tan(x)}(sec^{2}(x)ln(x)+(tan(x))/x)#.

Now let #z=cos(x^{tan(x)})=cos(y)# and compute, with the Chain Rule, #dz/dx=-sin(y) dy/dx#

#=-sin(x^{tan(x)})x^{tan(x)}(sec^{2}(x)ln(x)+(tan(x))/x)#

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Answer 2

Scarlett Allison

The derivative of cos(x)^tan(x) is given by the chain rule:

d/dx [cos(x)^tan(x)] = -sin(x)^(tan(x)+1) * (ln(cos(x))*tan(x) + sec(x)/cos(x))

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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.