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

1 Answer
Apr 6, 2018

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

Explanation:

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

and

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

First let's simplify noting that

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