Question

find the derivative of
$y = \frac{e^{7x}}{x^{7}+1}$
$\frac{dy}{dx}=$

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, $u = e^{7x}$ and $v=x^{7}+1$.

Step2: Find $u'$

Using the chain - rule, if $u = e^{7x}$, then $u'=7e^{7x}$.

Step3: Find $v'$

If $v=x^{7}+1$, then $v' = 7x^{6}$.

Step4: Substitute $u$, $u'$, $v$, $v'$ into quotient - rule

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

Answer:

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