How do you differentiate $3sin^5(2x)$ using the chain rule?

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Answer 1 · 2015-12-30T05:40:20

$30sin^4(2x)cos(2x)$

The first issue is that the sine function is to the fifth power.

According to the chain rule, $d/dx(u^5)=5u^4*(du)/dx$ .

So,

$d/dx(3sin^5(2x))=5(3sin^4(2x))d/dx(sin(2x))$

$=>15sin^4(2x)*d/dx(sin(2x))$

To find $d/dx(sin(2x))$ , use the chain rule again:

$d/dx(sin(u))=cos(u)*(du)/dx$

Thus,

$d/dx(sin(2x))=cos(2x)*d/dx(2x)=2cos(2x)$

Plug this back in:

$15sin^4(2x)*d/dx(sin(2x))$

$=>15sin^4(2x)*2cos(2x)$

$=>30sin^4(2x)cos(2x)$

How do you differentiate 3sin^5(2x) 3sin^5(2x) using the chain rule?