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UPSC NDA I and II 2020 Math Solved Paper

Question : 109 of 120

Marks: +1, -0

\tan^{-1} x

\cot^{-1} x

Solution:

Concept:

The derivative of one inverse trigonometric function with respect to another is found using the parametric differentiation formula:

\frac{du}{dv} = \frac{du/dx}{dv/dx}

Explanation:

Let

u = \tan^{-1} x

and

v = \cot^{-1} x

Differentiate both with respect to

x

\frac{du}{dx} = \frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^{2}}

\frac{dv}{dx} = \frac{d}{dx}(\cot^{-1}x) = \frac{-1}{1+x^{2}}

Now, use the formula

\frac{d\tan^{-1}x}{d\cot^{-1}x} = \frac{du}{dv} = \frac{du/dx}{dv/dx}

Substitute the derivatives:

\frac{du}{dv} = \frac{\frac{1}{1+x^{2}}}{\frac{-1}{1+x^{2}}} = -1

Answer:

The derivative is

-1

, which corresponds to option A.

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