# How do you differentiate y=e^x/x^7?

Jul 30, 2016

${y}^{'} = \frac{{e}^{x} \left(x - 7\right)}{x} ^ 8$

#### Explanation:

The easiest way, for me, is to first write this not as a quotient:

$y = {e}^{x} / {x}^{7} = {e}^{x} {x}^{-} 7$

From here, use the product rule, which states that if $y = f \left(x\right) g \left(x\right)$, then ${y}^{'} = {f}^{'} \left(x\right) g \left(x\right) + f \left(x\right) {g}^{'} \left(x\right)$.

So here, we see that $f \left(x\right) = {e}^{x}$, so ${f}^{'} \left(x\right) = {e}^{x}$ as well, and $g \left(x\right) = {x}^{-} 7$, so ${g}^{'} \left(x\right) = - 7 {x}^{-} 8$.

Thus:

${y}^{'} = {e}^{x} {x}^{-} 7 + {e}^{x} \left(- 7 {x}^{-} 8\right)$

Simplifying:

${y}^{'} = {e}^{x} / {x}^{7} - \frac{7 {e}^{x}}{x} ^ 8$

Common denominator:

${y}^{'} = \frac{x {e}^{x} - 7 {e}^{x}}{x} ^ 8$

${y}^{'} = \frac{{e}^{x} \left(x - 7\right)}{x} ^ 8$

Note that this can also be done with the quotient rule, which states that if $y = f \frac{x}{g} \left(x\right)$ then ${y}^{'} = \frac{{f}^{'} \left(x\right) g \left(x\right) - f \left(x\right) {g}^{'} \left(x\right)}{g \left(x\right)} ^ 2$.

So, in this case $f \left(x\right) = {e}^{x}$ so again ${f}^{'} \left(x\right) = {e}^{x}$, but $g \left(x\right) = {x}^{7}$ so ${g}^{'} \left(x\right) = 7 {x}^{6}$.

Thus:

${y}^{'} = \frac{{e}^{x} {x}^{7} - {e}^{x} \left(7 {x}^{6}\right)}{{x}^{7}} ^ 2 = \frac{{e}^{x} {x}^{6} \left(x - 7\right)}{x} ^ 14 = \frac{{e}^{x} \left(x - 7\right)}{x} ^ 8$