Question

1. find the derivative.

a) $y = xsin^{-1}x+ln(1 - x^{2})$

Explanation:

Step1: Apply product - rule and chain - rule

The derivative of a sum is the sum of derivatives. Let \(u = x\sin^{-1}x\) and \(v=\ln(1 - x^{2})\). First, find the derivative of \(u\) using the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = x\) and \(v=\sin^{-1}x\). The derivative of \(x\) is \(1\), and the derivative of \(\sin^{-1}x\) is \(\frac{1}{\sqrt{1 - x^{2}}}\). So, \(u^\prime=1\times\sin^{-1}x+x\times\frac{1}{\sqrt{1 - x^{2}}}=\sin^{-1}x+\frac{x}{\sqrt{1 - x^{2}}}\).

Step2: Find derivative of \(v\)

For \(v = \ln(1 - x^{2})\), use the chain - rule. Let \(t = 1 - x^{2}\), then \(v=\ln(t)\). The derivative of \(\ln(t)\) with respect to \(t\) is \(\frac{1}{t}\), and the derivative of \(t = 1 - x^{2}\) with respect to \(x\) is \(-2x\). So, \(v^\prime=\frac{-2x}{1 - x^{2}}\).

Step3: Find \(y^\prime\)

Since \(y = u + v\), then \(y^\prime=u^\prime+v^\prime\). Substitute \(u^\prime\) and \(v^\prime\) into the formula: \(y^\prime=\sin^{-1}x+\frac{x}{\sqrt{1 - x^{2}}}+\frac{-2x}{1 - x^{2}}\).

Answer:

\(y^\prime=\sin^{-1}x+\frac{x}{\sqrt{1 - x^{2}}}-\frac{2x}{1 - x^{2}}\)