Question

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)`