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)#