Question

How do you differentiate `f(x)=e^(cotsqrtx)` using the chain rule.?

Answer 1

`f'(x)=(-csc^2sqrtx*e^cotsqrtx)/(2sqrtx)`

Explanation:

First Issue: The `e` power. According to the chain rule, `d/dx(e^u)=e^u*u'`. Thus,

`f'(x)=e^cotsqrtx*d/dx(cotsqrtx)`

Second Issue: the cotangent function. Again through the chain rule, `d/dx(cotu)=-csc^2u*u'`. Thus,

`f'(x)=e^cotsqrtx*-csc^2sqrtx*d/dx(sqrtx)`

Now, to differentiate `sqrtx`, treat it as `x^(1/2)`. Thus, its derivative is `1/2x^(-1/2)` or `1/(2sqrtx)`.

`f'(x)=(-csc^2sqrtx*e^cotsqrtx)/(2sqrtx)`