Question

if f(x)=(sin^(- 1)(6x + 1)), then f(x)= note: the inverse of sin(x) can be entered as arcsin(x) or asin(x) attempt 2: 1 attempt remaining. submit answer next item

Explanation:

Step1: Apply chain - rule

The chain - rule states that if $y = u^n$ and $u = g(x)$, then $\frac{dy}{dx}=n\cdot u^{n - 1}\cdot\frac{du}{dx}$. Here $y = (\sin^{-1}(6x + 1))^7$, so let $u=\sin^{-1}(6x + 1)$ and $n = 7$. First, find the derivative of $y$ with respect to $u$: $\frac{dy}{du}=7u^{6}=7(\sin^{-1}(6x + 1))^{6}$.

Step2: Find the derivative of $u=\sin^{-1}(6x + 1)$ with respect to $x$

The derivative of $\sin^{-1}(t)$ with respect to $t$ is $\frac{1}{\sqrt{1 - t^{2}}}$. Using the chain - rule again, if $t = 6x+1$, then $\frac{du}{dx}=\frac{6}{\sqrt{1-(6x + 1)^{2}}}$.

Step3: Calculate $f'(x)$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. So $f'(x)=7(\sin^{-1}(6x + 1))^{6}\cdot\frac{6}{\sqrt{1-(6x + 1)^{2}}}=\frac{42(\sin^{-1}(6x + 1))^{6}}{\sqrt{1-(6x + 1)^{2}}}$.

Answer:

$\frac{42(\sin^{-1}(6x + 1))^{6}}{\sqrt{1-(6x + 1)^{2}}}$