find the derivative of the function. y = sin^(-1)(4x + 1) y =

Explanation:

Step1: Recall the chain - rule

If $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Let $u = 4x + 1$, so $y=\sin^{-1}(u)$.

Step2: Find the derivative of $y$ with respect to $u$

The derivative of $y = \sin^{-1}(u)$ with respect to $u$ is $y'_u=\frac{1}{\sqrt{1 - u^{2}}}$.

Step3: Find the derivative of $u$ with respect to $x$

The derivative of $u = 4x+1$ with respect to $x$ is $u'_x = 4$.

Step4: Apply the chain - rule

By the chain - rule $y'=y'_u\cdot u'_x$. Substitute $y'_u=\frac{1}{\sqrt{1 - u^{2}}}$ and $u'_x = 4$ into the formula, and replace $u$ with $4x + 1$. We get $y'=\frac{4}{\sqrt{1-(4x + 1)^{2}}}$.

Answer:

$\frac{4}{\sqrt{1-(4x + 1)^{2}}}$