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In Exercises \(7-12,\) use differentiation to verify the antiderivative formula. $$\int \frac{1}{\sqrt{1-u^{2}}} d u=\sin ^{-1} u+C$$

Short Answer

Expert verified
Yes, the derivative of \(\sin^{-1}u + C\) is \(\frac{1}{\sqrt{1-u^2}}\), and this verifies the given antiderivative formula.

Step by step solution

01

Understand the task

First, it's important to remember that if \(F(u)\) is an antiderivative of \(f(u)\), then the derivative of \(F(u)\) should equal \(f(u)\). Thus if we can show that the derivative of \(\sin^{-1}u + C\), where C is constant, equals to \(\frac{1}{\sqrt{1-u^2}}\), we will have verified the given antiderivative formula.
02

Differentiate the function

The derivative of \(\sin^{-1}u\) by the chain rule is \(\frac{1}{\sqrt{1-u^2}}\). And the derivative of any constant (in our case \(C\)) is zero. Hence, overall, the derivative of \(\sin^{-1}u + C\) is indeed \(\frac{1}{\sqrt{1-u^2}}\).

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