Hmm...
Let's let g(x)=y
We now apply this to f(x).
=>f(y)=sin(y) Substitute y with (x+3)^3
=>f(g(x))=sin((x+3)^3)
If you meant to find just f'(g(x))...
Just find the derivative of sin(y)
Since derivative of sin(x) is cos(x)...
=>f'(g(x))=cos((x+3)^3)
If you meant to ask for d/dx[f(g(x))], you use the chain rule:
d/dx[f(g(x))]=f'(g(x))*g'(x)
We already know that f'(g(x))=cos((x+3)^3)
=>d/dx[f(g(x))]=cos((x+3)^3)*d/dx[(x+3)^3]
We use the power rule:
d/dx[x^n]=nx^(n-1) if n is a constant.
=>d/dx[f(g(x))]=cos((x+3)^3)*3(x+3)^(3-1)*d/dx[x^1+3x^0]
Note here that the chain rule is applied again.
=>d/dx[f(g(x))]=cos((x+3)^3)*3(x+3)^2*1*x^(1-1)+3*0*x^(0-1)
=>d/dx[f(g(x))]=cos((x+3)^3)*3(x+3)^2*1*1+0
=>d/dx[f(g(x))]=cos((x+3)^3)*3(x+3)^2
That is the entire derivative calculated.
