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Differentiation

Section: Mathematics

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Question : 2 of 34

Marks: +1, -0

If

y = \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right) + \tan^{-1} \left( \frac{7x}{1 - 12x^2} \right)

x = 0, \; \frac{dy}{dx} =

[AP EAPCET 24 May 2025 S1]

6
7
9
10

Solution:

Given,

y = \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right) + \tan^{-1} \left( \frac{7x}{1 - 12x^2} \right)

Let

x = \tan \theta

\text{So},\; = \tan^{-1} \left( \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} \right) + \tan^{-1} \left( \frac{7\tan\theta}{1 - 12\tan^2\theta} \right)

\; = 3 \tan^{-1} x + \tan^{-1} \left( \frac{7x}{1 - 12x^2} \right)

Now, differentiate

y

w.r.t

x

, we get

\; \frac{dy}{dx} \; = 3 \cdot \frac{1}{1+x^2} + \frac{1}{1+\left(\frac{7x}{1-12x^2}\right)^2} \cdot \frac{d}{dx} \left( \frac{7x}{1-12x^2} \right)

\;=\; \frac{3}{1+x^2} + \frac{1}{1 + \frac{49x^2}{(1-12x^2)^2}} \cdot \frac{(1-12x^2) \cdot 7 - 7x(-24x)}{(1-12x^2)^2}

\;=\; \frac{3}{1+x^2} + \frac{(1-12x^2)^2}{(1-12x^2)^2 + 49x^2} \cdot \frac{7 - 84x^2 + 168x^2}{(1-12x^2)^2}

\;=\; \frac{3}{1+x^2} + \frac{7+84x^2}{(1-12x^2)^2 + 49x^2}

Now, At

x=0

\; \frac{dy}{dx} \Big|_{x=0} = \frac{3}{1+0} + \frac{7+84(0)}{(1-12(0)^2)^2 + 49.0^2} = 3 + \frac{7}{1+0} = 3+7 = 10

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