1. find the derivative.\na) $y = xsin^{-1}x+ln(1 - x^{2})$
1. find the derivative.\na) $y = xsin^{-1}x+ln(1 - x^{2})$
Answer
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}})