Remember the Chain Rule for Derivatives:
color(white)("XXX")color(red)((d f(g(x)))/(d x))XXXdf(g(x))dx=color(blue)((d f(g(x)))/(d g(x)) * (d g(x))/(dx)df(g(x))dg(x)⋅dg(x)dx
If we let
color(white)("XXX")color(brown)(f(x)=sin(x))XXXf(x)=sin(x) and
color(white)("XXX")color(magenta)(g(x)=5x)XXXg(x)=5x
then f(g(x))=sin(5x)color(white)("XX")f(g(x))=sin(5x)XX (the expression given in the question)
It is sometimes easier to replace g(x)g(x) with a simple variable.
Let's do that and let u=g(x)u=g(x)
So
color(white)("XXX"color(brown)()(d f(g(x)))/(d g(x)))XXXdf(g(x))dg(x) becomes color(brown)((d f(u))/(du)=(d sin(u))/(d u))df(u)du=dsin(u)du
Hopefully, you remember the basic trigonometric derivative:
color(white)("XXX")color(brown)((d sin(u))/(d u)=cos(u)=cos(g(x))=cos(5x))XXXdsin(u)du=cos(u)=cos(g(x))=cos(5x)
Also, since color(magenta)(g(x)=5x)g(x)=5x
color(white)("XXX")color(magenta)((d (g(x)))/(d x)=(d 5x)/(d x) =5)XXXd(g(x))dx=d5xdx=5
Therefore:
color(white)("XXX")color(red)((d f(g(x)))/(d x))XXXdf(g(x))dx=color(blue)((d f(g(x)))/(d g(x)) * (d g(x))/(dx)=color(magenta)(5) * color(brown)(cos(5x))df(g(x))dg(x)⋅dg(x)dx=5⋅cos(5x)
