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.
