How do you differentiate $f(x)=cot(5x)+csc(5x)$ ?

Question

3240 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-02T04:11:32

$f'(x) = -5csc(5x)(csc(5x)+cot(5x))$

$f(x) = cot(5x) + csc(5x)$

$f'(x) = d/dx(cot(5x)) +d/dx(csc(5x))$

Applying standard differentials and the chain rule to each term:

$f'(x )$ = $-csc^2 (5x) * d/dx(5x)$ $- cot(5x) * csc(5x)$ * $d/dx(5x)$

$= -csc^2 (5x) * 5 - cot(5x) * csc(5x) * 5$

$=-5csc(5x)(csc(5x)+cot(5x))$

How do you differentiate f(x)=cot(5x)+csc(5x)f(x)=cot(5x)+csc(5x)?