# Quick Chain Rule Question

• Feb 20th 2006, 01:15 PM
cinder
Quick Chain Rule Question
I have a homework problem similar to this, but I was unsure about the example in the book:

Find $\displaystyle F'(z)$ if $\displaystyle F(z) = (2z + 5)^3(3z - 1)^4$

The solution steps are as follows:

$\displaystyle F'(z) = (2z + 5)^3 \times 4(3z - 1)^3(3) + (3z - 1)^4 \times 3(2z + 5)^2(2)$
$\displaystyle = 6(2z + 5)^2(3z - 1)^3[2(2z + 5) + (3z - 1)]$
$\displaystyle = 6(2z + 5)^2(3z - 1)^3(7z + 9)$

I get a little lost during the second step, and where did that 6 come from?
• Feb 20th 2006, 01:42 PM
cinder
I think I figured it out. It looks like they factored, which I didn't do, but it gives the same answer.
• Feb 20th 2006, 02:24 PM
ThePerfectHacker
$\displaystyle F(z) = (2z + 5)^3(3z - 1)^4$
This uses the Product and Chain Rule.
Notice the derivative of $\displaystyle (2z+5)^3$ is $\displaystyle 6(2z+5)^2$. Now you get the six because your bring down the exponent (which is three) and multiply by the derivative of the inside (which is two) thus you get 6.
Similarly the derivative of $\displaystyle (3z-1)^4$ is $\displaystyle 12(3z-1)^3$.

Now by the product rule,
$\displaystyle u'v+uv'$ and already found $\displaystyle u',v'$ in the last paragraph. Thus,
$\displaystyle 6(2z+5)^2(3z-1)^4+12(2z+5)^3(3z-1)^3$
Thus, factor,
$\displaystyle 6(2z+5)^2(3z-1)^3((3z-1)+2(2z+5))$
Which becomes,
$\displaystyle 6(2z+5)^2(3z-1)^3(3z-1+4z+10)$
Which becomes,
$\displaystyle 6(2z+5)^2(3z-1)^3(7z+9)$
Q.E.D.