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#