Question

find the derivative of the algebraic function.
f(x)=\frac{x^{2}+13x + 36}{x^{2}-81}
f(x)=square

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x^{2}+13x + 36$, so $u^\prime=2x + 13$, and $v=x^{2}-81$, so $v^\prime = 2x$.

Step2: Apply quotient - rule

$f^\prime(x)=\frac{(2x + 13)(x^{2}-81)- (x^{2}+13x + 36)\times(2x)}{(x^{2}-81)^{2}}$.

Step3: Expand numerator

Expand $(2x + 13)(x^{2}-81)=2x\times x^{2}-2x\times81+13\times x^{2}-13\times81=2x^{3}-162x + 13x^{2}-1053$.
Expand $(x^{2}+13x + 36)\times(2x)=2x^{3}+26x^{2}+72x$.
Then the numerator is $(2x^{3}-162x + 13x^{2}-1053)-(2x^{3}+26x^{2}+72x)$.
$=2x^{3}-162x + 13x^{2}-1053 - 2x^{3}-26x^{2}-72x$.
$=- 13x^{2}-234x - 1053$.

Answer:

$\frac{-13x^{2}-234x - 1053}{(x^{2}-81)^{2}}$