Question

question
determine $t(x)$ if $t(x)=\frac{e^{x}}{-8x^{7}}$.
select the correct answer below:
$t(x)=\frac{e^{x - 1}(7e + x^{2})}{-8x^{8}}$
$t(x)=\frac{e^{x}(-x - 7)}{-8x^{8}}$
$t(x)=\frac{e^{x}(x - 7)}{-8x^{8}}$
$t(x)=\frac{e^{x}}{-56x^{6}}$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $t(x)=\frac{u(x)}{v(x)}$, then $t'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}$. Here, $u(x)=e^{x}$, $u'(x)=e^{x}$, $v(x)= - 8x^{7}$, and $v'(x)=-56x^{6}$.

Step2: Substitute values

$t'(x)=\frac{e^{x}(-8x^{7})-e^{x}(-56x^{6})}{(-8x^{7})^{2}}=\frac{e^{x}(-8x^{7}+56x^{6})}{64x^{14}}=\frac{e^{x}x^{6}(-8x + 56)}{64x^{14}}=\frac{e^{x}(-x + 7)}{-8x^{8}}=\frac{e^{x}(x - 7)}{-8x^{8}}$

Answer:

$t'(x)=\frac{e^{x}(x - 7)}{-8x^{8}}$ (corresponding to the third option)