Question

evaluate the derivative of the following function.
f(x)=5x sin^{-1}x
f(x)=\square

Explanation:

Step1: Apply the product rule

The product rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = 5x\) and \(v=\sin^{-1}x\). First, find \(u^\prime\) and \(v^\prime\).
\(u^\prime=\frac{d(5x)}{dx}=5\).
The derivative of \(y = \sin^{-1}x\) is \(y^\prime=\frac{1}{\sqrt{1 - x^{2}}}\), so \(v^\prime=\frac{1}{\sqrt{1 - x^{2}}}\).

Step2: Substitute into the product rule formula

\(f^\prime(x)=u^\prime v+uv^\prime\)
\(f^\prime(x)=5\cdot\sin^{-1}x+5x\cdot\frac{1}{\sqrt{1 - x^{2}}}\)

Answer:

\(5\sin^{-1}x+\frac{5x}{\sqrt{1 - x^{2}}}\)