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

$-cotx/sinx=-cscxcotx.$

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