Question

differentiate
$\frac{ {x}^{2} - 1 }{ {x}^{3} + 1 }$
by differentiating with respect to x​

Answer 1

Answer:

The required answer is  $\frac{3x^{2} -4x+2}{(x^{2} -x+1)^{2}}$

Step-by-step explanation:

 $\frac{d}{dx}$ (  $\frac{x^{2}-1 }{x^{3}+1}$ )

To make the calculations easier, we simplify the numerator and the denominator.

 $\frac{d}{dx}$ (  $\frac{x^{2}-1 }{x^{3}+1}$ ) =  

Cancelling (x + 1) we get

$\frac{d}{dx}$ $\frac{(x-1)}{(x^{2} -x +1)}$

We know that $\frac{d}{dx}$ $\frac{Numerator}{Denominator}$ = $\frac{D*\frac{d}{dx} N + N*\frac{d}{dx} D}{D^{2} }$

where N is Numerator and D is Denominator

Solving the given question using this formula, we get

$\frac{(x^{2}-x+1) (1-0) + (x-1) (2x-1+0) }{(x^{2} -x+1)^{2}}$

= $\frac{x^{2} -x +1 +2x^{2} -x-2x+1}{(x^{2} -x+1)^{2}}$

= $\frac{3x^{2} -4x+2}{(x^{2} -x+1)^{2}}$

Which is the required solution.