#(d sin(5x))/(dx)=color(green)(5cos(5x))#

Remember the Chain Rule for Derivatives:
#color(white)("XXX")color(red)((d f(g(x)))/(d x))#=#color(blue)((d f(g(x)))/(d g(x)) * (d g(x))/(dx)#

If we let
#color(white)("XXX")color(brown)(f(x)=sin(x))# and
#color(white)("XXX")color(magenta)(g(x)=5x)#
then #f(g(x))=sin(5x)color(white)("XX")# (the expression given in the question)

It is sometimes easier to replace #g(x)# with a simple variable.
Let's do that and let #u=g(x)#
So
#color(white)("XXX"color(brown)()(d f(g(x)))/(d g(x)))# becomes #color(brown)((d f(u))/(du)=(d sin(u))/(d u))#

Hopefully, you remember the basic trigonometric derivative:
#color(white)("XXX")color(brown)((d sin(u))/(d u)=cos(u)=cos(g(x))=cos(5x))#

Also, since #color(magenta)(g(x)=5x)#
#color(white)("XXX")color(magenta)((d (g(x)))/(d x)=(d 5x)/(d x) =5)#

Therefore:
#color(white)("XXX")color(red)((d f(g(x)))/(d x))#=#color(blue)((d f(g(x)))/(d g(x)) * (d g(x))/(dx)=color(magenta)(5) * color(brown)(cos(5x))#