Question:

If $y = \cot^{-1} \, (x^2)$, then the value of $ \frac{dy}{dx}$ is equal to

Updated On: Nov 4, 2023

$\frac{2x}{1 + x^4}$
$\frac{2x}{\sqrt{1 + 4x}}$
$\frac{-2x}{1 + x^4}$
$\frac{- 2x}{\sqrt{1 + x^2}}$

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The Correct Option is C

Solution and Explanation

Let $y =\cot^{-1} (x^2)$ $\Rightarrow \:\:\: \cot \, y = x^2$ Diff both side, w.r.t. $'x' - cosec^2 y. \frac{dy}{dx} = 2x$ $\frac{dy}{dx} = \frac{2x}{-cosec^2 y}$ $ =\frac{2x}{-(1+ \cot^2 y)} = \frac{2x}{-(x^4 + 1)} = \frac{-2x}{(x^4 + 1)}$

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Definition of Differentiability

f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by

Definition of Continuity

Mathematically, a function is said to be continuous at a point x = a, if

It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.

If the function is unspecified or does not exist, then we say that the function is discontinuous.

If $y = \cot^{-1} \, (x^2)$, then the value of $ \frac{dy}{dx}$ is equal to

The Correct Option is C

Solution and Explanation

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