differentiate.\ny = \\frac{7x}{e^{x}}\ny = 7 - e^{x} ×\nresources\nread it watch it

Explanation:

Step1: Recall quotient - rule

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

Step2: Find $u'$ and $v'$

The derivative of $u = 7x$ with respect to $x$ is $u'=7$ (since the derivative of $ax$ with $a = 7$ is $a$). The derivative of $v = e^{x}$ with respect to $x$ is $v'=e^{x}$ (since the derivative of $e^{x}$ is $e^{x}$).

Step3: Apply the quotient - rule

Substitute $u$, $v$, $u'$, and $v'$ into the quotient - rule formula: [ \begin{align*} y'&=\frac{7\cdot e^{x}-7x\cdot e^{x}}{(e^{x})^{2}}\ &=\frac{7e^{x}(1 - x)}{e^{2x}}\ &=\frac{7(1 - x)}{e^{x}} \end{align*} ]

Answer:

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