How do you differentiate $f(x)=csc5x^5$ ?

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Answer 1 · 2017-07-25T07:38:16

$f'(x)=-25x^4cot5x^5csc5x^5$

we will need to use the chain rule;

$(dy)/(dx)=(dy)/(du)xx(du)/(dx)$

let

$y=csc5x^5$

let $" "u=5x^5$

$:.y=cscu$

$=>(dy)/(du)=-cotucscu$

$(du)/(dx)=25x^4$

$(dy)/(dx)=(dy)/(du)xx(du)/(dx)$

$=>(dy)/(dx)=(-cotucscu)(25x^4)$

substituting back for $u$ and#f tidying up

$f'(x)=-25x^4cot5x^5csc5x^5$

How do you differentiate f(x)=csc5x^5f(x)=csc5x^5?