Question

What is the derivative of `f(x)=(cos x*cscx) / (1+ tan^2 x)`?

Answer 1

`f'(x)=-2cos^2(x)-cot^2(x)`

Explanation:

First, simplify the expression by converting everything to `sin` and `cos`.

`f(x)=frac{cos(x)*frac{1}{sin(x)}}{frac{cos^2(x)}{cos^2(x)}+frac{sin^2(x)}{cos^2(x)}}`

`=frac{cos^3(x)}{sin(x)}`

`f'(x)=frac{d}{dx}(frac{cos^3(x)}{sin(x)})`

`=frac{sin(x)frac{d}{dx}(cos^3(x))-cos^3(x)frac{d}{dx}(sin(x))}{sin(x)^2}`

`=frac{sin(x)(3cos^2(x)(-sin(x)))-cos^3(x)(cos(x))}{sin^2(x)}`

`=frac{-3sin^2(x)cos^2(x)-cos^4(x)}{sin^2(x)}`

`=frac{-3sin^2(x)cos^2(x)+cos^2(x)(sin^2(x)-1)}{sin^2(x)}`

`=-3cos^2(x)+cos^2(x)-cot^2(x)`

`=-2cos^2(x)-cot^2(x)`