If #f(x) =xe^(5x+4) # and #g(x) = cos2x #, what is #f'(g(x)) #?

whilst the intention of this question may have been to encourage the use of the chain rule on both #f(x)# and #g(x)# - hence, why this is filed under Chain Rule - that is not what the notation asks for.

to make the point we look at the definition

#f'(u) = (f(u + h) - f(u))/(h) #

or

#f'(u(x)) = (f(u(x) + h) - f(u(x)))/(h) #

the prime means differentiate wrt to whatever is in the brackets

here that means, in Liebnitz notation: #(d(f(x)))/(d(g(x)) #

contrast with this the full chain rule description:

# (f\circ g)'(x) = f'(g(x))\cdot g'(x) #

So, in this case, #u = u(x) = cos 2x# and so the notation requires simply the derivative of #f(u) # wrt to #u#, and then with #x to cos 2x #, ie #cos 2x# inserted as x in the resultant derivative

So here
# f'(cos 2x) qquad["let " u = cos 2x#]#

#= f'(u)#

by the product rule
# = (u)' e^(5u + 4) + u ( e^(5u + 4) )' #

#= e^(5u + 4) + u *5 e^(5u + 4) #

#= e^(5u + 4) (1 + 5u) #

So
#f'(g(x))= #f'(cos 2x) #

#= e^(5cos 2x + 4) (1 + 5cos 2x) #

in short

#f'(g(x)) ne (f\circ g)'(x) #

#f(x)=xe^(5x+4)#
To find #f'(g(x))#, first we have to find #f'(x)# then we have to substitute #x# by #g(x)#

#f'(x)=e^(5x+4)+5xe^(5x+4)#
#f'(x)=e^(5x+4)(1+5x)#
Let us substitute #x# by #f(x)#
#f'(g(x))=e^(5cos(2x)+4)(1+5cos2x)#