## First step to finding the derivative of 20x/(x^2-4)^2

Limits, differentiation, related rates, integration, trig integrals, etc.

### First step to finding the derivative of 20x/(x^2-4)^2

I got f'(x)=[(x^2-4)^2(20)-2(x^2-4)(2x)(20x)]/(x^2-4)^4
=(-80x^4+340x^2-80)/(x^2-4)^3
and simplify

But I am pretty sure my first step was wrong, like I did the chain or product rule wrong because my answer does not match up to what my TI-89 says it is supposed to be.
Cafe au lait

Posts: 6
Joined: Mon Jan 03, 2011 11:02 pm

### Re: First step to finding the derivative of 20x/(x^2-4)^2

Cafe au lait wrote:I got f'(x)=[(x^2-4)^2(20)-2(x^2-4)(2x)(20x)]/(x^2-4)^4
=(-80x^4+340x^2-80)/(x^2-4)^3
and simplify

But I am pretty sure my first step was wrong, like I did the chain or product rule wrong because my answer does not match up to what my TI-89 says it is supposed to be.

$\frac{(x^2-4)^2(20)-2(x^2-4)(2x)(20x)}{(x^2-4)^4}$

$\Rightarrow\frac{20(x^2-4)\left[(x^2-4)-2(2x)(x)\right]}{(x^2-4)^4}$

$\Rightarrow\frac{20\left[x^2-4-4x^2\right]}{(x^2-4)^3}$

$\Rightarrow\frac{20\left[-3x^2-4\right]}{(x^2-4)^3}$

$\Rightarrow\frac{-20\left[3x^2+4\right]}{(x^2-4)^3}$

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

### Re: First step to finding the derivative of 20x/(x^2-4)^2

Cafe au lait

Posts: 6
Joined: Mon Jan 03, 2011 11:02 pm

### Re: First step to finding the derivative of 20x/(x^2-4)^2

Cafe au lait wrote:... to what my TI-89 says it is supposed to be.

Cafe au lait wrote:That's the answer my calculator gave me! ...

I know. You are not the only one with a ti-89

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA