# Math Help - [SOLVED] quotient rule.

1. ## [SOLVED] quotient rule.

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

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

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

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

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

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

????

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

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

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

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

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

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

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