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The derivative of $\sin^{-1}x$ is
A$\dfrac{1}{1+x^2}$
B$-\dfrac{1}{\sqrt{1-x^2}}$
C$\dfrac{1}{\sqrt{1-x^2}}$
D$\dfrac{1}{\sqrt{1+x^2}}$
Answer & Solution
Correct answer: C. $\dfrac{1}{\sqrt{1-x^2}}$
This is a standard inverse trigonometric derivative: $\dfrac{d}{dx}(\sin^{-1}x)=\dfrac{1}{\sqrt{1-x^2}}$.
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