Question

1. given $y = \frac{x^{7}-1}{x^{7}+1}$, find $\frac{dy}{dx}$. $\frac{dy}{dx}=$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Here, $u = x^{7}-1$, $u'=7x^{6}$, $v = x^{7}+1$, and $v'=7x^{6}$.

Step2: Substitute into quotient - rule formula

$\frac{dy}{dx}=\frac{7x^{6}(x^{7}+1)-(x^{7}-1)\times7x^{6}}{(x^{7}+1)^{2}}$.

Step3: Expand the numerator

$\frac{dy}{dx}=\frac{7x^{13}+7x^{6}-(7x^{13}-7x^{6})}{(x^{7}+1)^{2}}=\frac{7x^{13}+7x^{6}-7x^{13}+7x^{6}}{(x^{7}+1)^{2}}$.

Step4: Simplify the numerator

$\frac{dy}{dx}=\frac{14x^{6}}{(x^{7}+1)^{2}}$.

Answer:

$\frac{14x^{6}}{(x^{7}+1)^{2}}$