How do you differentiate #[sin[(3x+7)^2]^3]#?

Question

1836 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-06T18:51:26

#(dsin{[(3x+7)^2]^3})/(dx)=18(3x+7)^5cos[(3x+7)^6]#

The notation on this problem is a little ambiguous. I am going to assume that

#sin[(3x+7)^2]^3=sin{[(3x+7)^2]^3}#

and

#sin[(3x+7)^2]^3ne{sin[(3x+7)^2]}^3#

First let's simplify noting that

#sin{[(3x+7)^2]^3}=sin[(3x+7)^6]#.

This is a job for the chain rule. In fact we apply it twice.

#(df(g(h(x))))/(dx)=f^'(g(h(x)))*g^'(h(x))*h^'(x)#

Here, #f(x)=sinx#, #g(x)=x^6#, and #h(x)=3x+7#. This means that #f^'(x)=cosx#, #g^'(x)=6x^5#, and #h^'(x)=3#.

So

#(dsin[(3x+7)^6])/(dx)=cos[(3x+7)^6]*6(3x+7)^5*3#

#=18(3x+7)^5cos[(3x+7)^6]#.