find the derivative of the function.\ny = 2e^{4x^{2}-3}\n$\frac{dy}{dx}=$

Explanation:

Step1: Recall the chain - rule

If $y = 2e^{u}$ and $u = 4x^{2}-3$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$.

Step2: Differentiate $y$ with respect to $u$

The derivative of $y = 2e^{u}$ with respect to $u$ is $\frac{dy}{du}=2e^{u}$.

Step3: Differentiate $u$ with respect to $x$

The derivative of $u = 4x^{2}-3$ with respect to $x$ is $\frac{du}{dx}=8x$.

Step4: Apply the chain - rule

$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=2e^{u}\cdot8x$. Substitute $u = 4x^{2}-3$ back in, we get $\frac{dy}{dx}=16xe^{4x^{2}-3}$.

Answer:

$16xe^{4x^{2}-3}$