Question

How do you find the derivative of `sinx^tanx`?

Answer 1

Use logarithmic differentiation to get `d/dx(sin(x)^{tan(x)}) = (1+ln(sin(x))sec^2(x))*sin(x)^{tan(x)}`.

Explanation:

First, let `y=sin(x)^{tan(x)}`.

Next, take the natural logarithm of both sides and use a property of logarithms to get `ln(y)=tan(x)ln(sin(x))`.

Next, differentiate both sides with respect to `x`, keeping in mind that `y` is a function of `x` and using the Chain Rule and Product Rule to get `1/y*dy/dx=sec^{2}(x)ln(sin(x))+tan(x)*1/(sin(x))*cos(x)`

`=1+ln(sin(x))sec^{2}(x)`.

Multiplying both sides by `y=sin(x)^{tan(x)}` now gives the final answer to be `d/dx(sin(x)^{tan(x)}) = (1+ln(sin(x))sec^2(x))*sin(x)^{tan(x)}`.