# Finding The Derivative of a composition of a function....

#### soulrain

so I know feed f(x) back into itself and differentiate and then plug 1 in for g prime but it gets super complex is there a trick I am missing?

#### MarkFL

Re: Finding The Derivative of a composition of a function...

Use the chain rule to state:

$$\displaystyle g'(x)=f'\(f(x)$$f'(x)\)

hence:

$$\displaystyle g'(1)=f'\(f(1)$$f'(1)\)

So, we need to compute, using the product and chain rules:

$$\displaystyle f'(x)=x\(e^{x^2}\cdot2x$$+(1)$$e^{x^2}$$=e^{x^2}$$2x^2+1$$\)

We now need:

$$\displaystyle f(1)=e$$

$$\displaystyle f'\(f(1)$$=f'(e)=e^{e^2}$$2e^2+1$$\)

$$\displaystyle f'(1)=3e$$

and so:

$$\displaystyle g'(1)=\(e^{e^2}\(2e^2+1$$\)$$3e$$=3e^{e^2+1}$$2e^2+1$$\)

#### soulrain

Re: Finding The Derivative of a composition of a function...

Thanks mark working through your post now thank you so much!!!

#### soulrain

Re: Finding The Derivative of a composition of a function...

not only worked through it but totally understand it...

I kept substituting directly instead of doing the chain rule on the general statement and it turned out no bueno.

#### MarkFL

Re: Finding The Derivative of a composition of a function...

Yeah, without the chain rule that would be a real mess to work through. :mrgreen:

#### soulrain

Re: Finding The Derivative of a composition of a function...

A real mess indeed!