Question

If $y = \tan^{-1}(\frac{3x - x^3}{1-3x^2}) + \tan^{-1}(\frac{7x}{1-12x^2})$, then at $x=0$, $\frac{dy}{dx} =$

• A. 6
• B. 7
• C. 9
• D. 10

Hint:

Chain rule of differentiation. At $x=0$, the derivative simplifies significantly.

Answer 1

The Correct Option is D

To find the derivative of \( y \) with respect to \( x \) at \( x = 0 \), where \( y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) + \tan^{-1}\left(\frac{7x}{1 - 12x^2}\right) \), we will use the derivatives of inverse tangent functions and sum rule of differentiation. The derivative of \( \tan^{-1}(u) \) with respect to \( x \) is \( \frac{1}{1+u^2}\cdot\frac{du}{dx} \).

Let \( u_1 = \frac{3x - x^3}{1 - 3x^2} \) and \( u_2 = \frac{7x}{1 - 12x^2} \). We differentiate each:

For \( u_1 \):
\(\frac{du_1}{dx} = \frac{(1-3x^2)(3 - 3x^2) - (3x-x^3)(-6x)}{(1-3x^2)^2}\)
\(= \frac{3 - 9x^2 + 9x^4 + 18x^2 - 6x^4}{(1-3x^2)^2}\)
\(= \frac{3 + 9x^2 + 3x^4}{(1-3x^2)^2}\)

Substituting back at \( x = 0 \), \(\frac{du_1}{dx}\Big|_{x=0} = \frac{3}{1} = 3\).

Thus, \(\frac{d}{dx}\left(\tan^{-1}(u_1)\right)\Big|_{x=0} = \frac{1}{1+(u_1{\big|_{x=0}})^2} \cdot \frac{du_1}{dx}\Big|_{x=0} = \frac{1}{1+(0)^2} \cdot 3 = 3\).

For \( u_2 \):
\(\frac{du_2}{dx} = \frac{(1-12x^2)(7) - (7x)(-24x)}{(1-12x^2)^2}\)
\(= \frac{7 - 84x^2 + 168x^2}{(1-12x^2)^2}\)
\(= \frac{7 + 84x^2}{(1-12x^2)^2}\)

Substituting back at \( x = 0 \), \(\frac{du_2}{dx}\Big|_{x=0} = \frac{7}{1} = 7\).

Thus, \(\frac{d}{dx}\left(\tan^{-1}(u_2)\right)\Big|_{x=0} = \frac{1}{1+(u_2{\big|_{x=0}})^2} \cdot \frac{du_2}{dx}\Big|_{x=0} = \frac{1}{1+(0)^2} \cdot 7 = 7\).

Then, combine derivatives for both inverse tangent functions using the sum rule:
\(\frac{dy}{dx}\Big|_{x=0} = 3 + 7 = 10\).

Therefore, the value of \(\frac{dy}{dx}\) at \( x = 0 \) is 10.