Question

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

Answer 1

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

Explanation:

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)`

Answer 2

`f'(g(x))=e^(5cos(2x)+4)(1+5cos2x)`

Explanation:

`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)`