Question

if f(x)=(sin⁻¹(x + 6))³, 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 \(\frac{dy}{dx}=n\cdot u^{n - 1}\cdot g^{\prime}(x)\). Here \(y = (\sin^{-1}(x + 6))^{3}\), let \(u=\sin^{-1}(x + 6)\), so \(y = u^{3}\). First, find \(\frac{dy}{du}\) and \(\frac{du}{dx}\).
The derivative of \(y = u^{3}\) with respect to \(u\) is \(\frac{dy}{du}=3u^{2}=3(\sin^{-1}(x + 6))^{2}\).
The derivative of \(u=\sin^{-1}(x + 6)\) with respect to \(x\) is \(\frac{du}{dx}=\frac{1}{\sqrt{1-(x + 6)^{2}}}\) (since the derivative of \(\sin^{-1}(t)\) with respect to \(t\) is \(\frac{1}{\sqrt{1 - t^{2}}}\), here \(t=x + 6\)).

Step2: Calculate \(f^{\prime}(x)\)

By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\), so \(f^{\prime}(x)=3(\sin^{-1}(x + 6))^{2}\cdot\frac{1}{\sqrt{1-(x + 6)^{2}}}\).

Answer:

\(\frac{3(\sin^{-1}(x + 6))^{2}}{\sqrt{1-(x + 6)^{2}}}\)