Question

for the function y = -2x^4+7xe^x, find \frac{dy}{dx}. answer: \frac{dy}{dx}=

Explanation:

Step1: Apply sum - difference rule

The derivative of a sum/difference of functions is the sum/difference of their derivatives. So, $\frac{dy}{dx}=\frac{d}{dx}(-2x^{4})+\frac{d}{dx}(7xe^{x})$.

Step2: Differentiate $-2x^{4}$

Using the power rule $\frac{d}{dx}(ax^{n}) = nax^{n - 1}$, for $a=-2$ and $n = 4$, we have $\frac{d}{dx}(-2x^{4})=-2\times4x^{4 - 1}=-8x^{3}$.

Step3: Differentiate $7xe^{x}$

Using the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u = 7x$ and $v=e^{x}$. $\frac{du}{dx}=7$ and $\frac{dv}{dx}=e^{x}$. Then $\frac{d}{dx}(7xe^{x})=7x\cdot e^{x}+7\cdot e^{x}=7e^{x}(x + 1)$.

Step4: Combine results

$\frac{dy}{dx}=-8x^{3}+7e^{x}(x + 1)$.

Answer:

$-8x^{3}+7e^{x}(x + 1)$