1. ## chain rule

find f '(x)

f(x)=4/(3xto2nd-2x+1)to3rd

2. Originally Posted by homerb
find f '(x)

f(x)=4/(3xto2nd-2x+1)to3rd
Note that $\frac{4}{(3x^2-2x+1)^3}=4(3x^2-2x+1)^{-3}$

Do you know how to apply the chain rule?

$\frac{\,d}{\,dx}\left[f(g(x))\right]=f'(g(x))\cdot g'(x)$

Let $f(x)=4x^{-3}$ and let $g(x)=3x^2-2x+1$

Can you take it from here?

--Chris

3. no I can't

4. $\frac{\,d}{\,dx}\left[f(g(x))\right]=f'(g(x))\cdot g'(x)$

Let $f(x)=4x^{-3}$ and let $g(x)=3x^2-2x+1$

Then we see that $f'(x)=-12x^{-4}$ and $g'(x)=6x-2$

So when we apply the chain rule, we see that $\frac{\,d}{\,dx}\left[4(3x^2-2x+1)^{-3}\right]=-12(3x^2-2x+1)^{-4}\cdot (6x-2)$

Can you take it from here and simplify this?

--Chris

5. I try to do it, but I can't. This is my first time doing this. If you can simplify this more

6. Originally Posted by homerb
I try to do it, but I can't. This is my first time doing this. If you can simplify this more
The farthest it can be simplified:

$-12(3x^2-2x+1)^{-4}\cdot (6x-2)=-\frac{24(3x-1)}{(3x^2-2x+1)^4}$

--Chris

7. homer you times the -12 by the 6x-2 for the numerator and the equation in the parenthesis goes in the denominator and it becomes a positive exponent because you put it in the denominator