Derivative with general exponential and logarithmic fuctions

[Beeorz]

Homework Statement

f(x)=(((x^7)+3)((x^3)+2)) / (((x+3)^2)((x^6)+5))
Find f'(x)

Homework Equations

d/dx lnx = 1/x
d/dx ln(f(x)) = f'(x)/f(x)

The Attempt at a Solution

ln(f(x)) = ln((x^7)+3) + ln((x^3)+2) - ln((x+3)^2) - ln ((x^6)+5)
f'(x)/f(x) = (7x^6)/((x^7)+3) + (3x^2)/((x^3)+2) - (2)/(x+3) - (6x^5)/((x^6)+5)

f'(x) = [(((x^7)+3)((x^3)+2)) / (((x+3)^2)((x^6)+5))]*[(7x^6)/((x^7)+3) + (3x^2)/((x^3)+2) - (2)/(x+3) - (6x^5)/((x^6)+5)]

My answer may be correct - just wanting someone to give it a quick check.

I know it looks terrible just to even look at. If someone could please show me where/how to use better syntax I would greatly appreciate it. I'll even rewrite the problem so that it'll look more friendly :P

 

[CompuChip]

Hmm I missed out on this thread before, but actually that's a nice trick that I hadn't heard / thought of before.
Actually this makes it very easy to prove that in general, the derivative of
$$\phi(x) = \frac{f g}{h k}$$
(with f, g, h, k functions of x) is
$$\phi' = \frac{f' g}{h k} + \frac{f g'}{h k} - \frac{f g}{h^2 k} h' - \frac{f g}{h k^2} k'$$

(And yes, your answer is correct)

 

[HallsofIvy]

It's just a fairly standard application of the logarithmic derivative. And yes, the answer is correct.