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

Jan 2012
159
0


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?
 
Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
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}\(2e^2+1\)\)
 
Jan 2012
159
0
Re: Finding The Derivative of a composition of a function...

Thanks mark working through your post now thank you so much!!!
 
Jan 2012
159
0
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.
 
Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
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:
 
Jan 2012
159
0
Re: Finding The Derivative of a composition of a function...

A real mess indeed!