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What is $\dfrac{d}{dx}(\tan^{-1}x)$?
A$\dfrac{1}{1-x^2}$
B$\dfrac{1}{\sqrt{1+x^2}}$
C$\dfrac{1}{1+x^2}$
D$-\dfrac{1}{1+x^2}$
Answer & Solution
Correct answer: C. $\dfrac{1}{1+x^2}$
The standard result is $\dfrac{d}{dx}(\tan^{-1}x)=\dfrac{1}{1+x^2}$. Option B is a common confusion with inverse trigonometric forms involving square roots.
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