How do you differentiate #f(x)=csc(2x-x^3) # using the chain rule?

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Answer 1 · 2016-01-19T01:28:18

#f'(x)=csc(2x-x^3)cot(2x-x^3)(3x^2-2)#

Using the chain rule :

#f'(x) = d/dxcsc(2x-x^3)#

#= -csc(2x-x^3)cot(2x-x^3)(d/dx2x-x^3)#

#= -csc(2x-x^3)cot(2x-x^3)((d/dx2x)-(d/dxx^3))#

#= -csc(2x-x^3)cot(2x-x^3)(2-3x^2)#

#=csc(2x-x^3)cot(2x-x^3)(3x^2-2)#