Question

find the derivative of the function.
y = 7e^{-x}+e^{3x}
\frac{dy}{dx}=\square

Explanation:

Step1: Recall derivative rules

The derivative of $e^{ax}$ is $ae^{ax}$ and the derivative of a sum of functions $u + v$ is $u'+v'$.

Step2: Differentiate $7e^{-x}$

Using the rule, if $u = 7e^{-x}$, then $u'=7\times(-1)e^{-x}=- 7e^{-x}$.

Step3: Differentiate $e^{3x}$

If $v = e^{3x}$, then $v' = 3e^{3x}$.

Step4: Find the derivative of $y$

Since $y=7e^{-x}+e^{3x}$, by the sum - rule of derivatives $\frac{dy}{dx}=u' + v'=-7e^{-x}+3e^{3x}$.

Answer:

$-7e^{-x}+3e^{3x}$