# Thread: [SOLVED] quotient rule.

1. ## [SOLVED] quotient rule.

$\displaystyle f(x)=\frac{10x+3}{5x^2+3x-6}$

$\displaystyle f(x)=\frac{[(10)(5x^2+3x-6)]-[10x+3)(10x+3]}{(5x^2+3x-6)}$

$\displaystyle f(x)=\frac{(50x^2+30x-60)-(10x+3)^2}{(5x^2+3x-6)^2}$

$\displaystyle f(x)=\frac{(50x^2+30x-60)-(10x^2+9)}{(5x^2+3x-6)^2}$

$\displaystyle f'(x)= \frac{40x^2+30x-69}{(5x^2+3x-6)^2}$

Did i do that right or is the end part
$\displaystyle \frac{40x^2+30x-51}{(5x^2+3x-6)^2}$

????

2. Originally Posted by el123
$\displaystyle f(x)=\frac{10x+3}{5x^2+3x-6}$

$\displaystyle f(x)=\frac{[(10)(5x^2+3x-6)]-[10x+3)(10x+3]}{(5x^2+3x-6)}$

$\displaystyle f(x)=\frac{(50x^2+30x-60)-(10x+3)^2}{(5x^2+3x-6)^2}$

$\displaystyle f(x)=\frac{(50x^2+30x-60)-(10x^2+9)}{(5x^2+3x-6)^2}$

$\displaystyle f'(x)= \frac{40x^2+30x-69}{(5x^2+3x-6)^2}$

Did i do that right or is the end part
$\displaystyle \frac{40x^2+30x-51}{(5x^2+3x-6)^2}$

????
Check your answer here: Wolfram|Alpha (click Show Steps).

3. On the fourth line you seem to have gone from $\displaystyle (10x + 3)^2$ to $\displaystyle 10x^2 + 9$ which is mentioned here:

Freshman's Dream - ProofWiki

as it's such a popular mistake to make.

4. Thats was all a bit confusing there for a while, but i have figured it out.

Thanks for both those sites! Im sure they will come in handy.