What is the derivative of #y=csc^3(x)#?

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Answer 1 · 2018-03-03T22:12:39

#y'=-3csc^3(x)cot(x)#

We can rewrite #y=csc^3(x)# as #y=(cscx)^3.#

Using the Power rule and chain rule, we get

#y'=3(cscx)^2*d/dx(cscx)=3csc^2(x)*-csc(x)cot(x)=-3csc^3(x)cot(x)#

As #d/dxcsc(x)=-csc(x)cot(x).#

Here's a proof of this derivative:

#csc(x)=1/sin(x)=(sinx)^-1#

Since #d/dxsin(x)=cos(x),# using the Power Rule and Chain Rule on #d/dx(sinx)^-1# yields

#d/dx(sinx)^-1=-(sinx)^-2cosx=-cosx/sin^2x=-cosx/(sinx*sinx)#

Recall that #cosx/sinx=cotx#

Thus, #-cosx/(sinx*sinx)=-cotx/sinx#. #1/sinx=cscx,# so

#-cotx/sinx=-cscxcotx.#