Question:
If
\[
y=\tan(3\tan^{-1}x),
\]
then
\[
(1-3x^2)\frac{d^2y}{dx^2}-12x\frac{dy}{dx}
\]
is equal to:
If
\[
y=\tan(3\tan^{-1}x),
\]
then
\[
(1-3x^2)\frac{d^2y}{dx^2}-12x\frac{dy}{dx}
\]
is equal to:
Show Hint
For expressions involving \(\tan(3\tan^{-1}x)\), first use the identity
\[
\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}.
\]
This makes differentiation easier.
Updated On: Jun 26, 2026
- \(6(x+y)\)
- \(6(y-x)\)
- \(6y\)
- \(-6x\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Simplify the given function.
Given, \[ y=\tan(3\tan^{-1}x). \] Let \[ \theta=\tan^{-1}x. \] Then, \[ x=\tan\theta. \] So, \[ y=\tan 3\theta. \] Using the formula, \[ \tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}, \] we get \[ y=\frac{3x-x^3}{1-3x^2}. \]
Step 2: Differentiate \(y\).
After differentiating, \[ \frac{dy}{dx}=\frac{3(1+x^2)^2}{(1-3x^2)^2}. \] Again differentiating, \[ \frac{d^2y}{dx^2} = \frac{12x(1+x^2)(1-3x^2)+36x(1+x^2)^2}{(1-3x^2)^3}. \]
Step 3: Substitute in the given expression.
Now, \[ (1-3x^2)\frac{d^2y}{dx^2}-12x\frac{dy}{dx} \] simplifies to \[ 6\left(\frac{3x-x^3}{1-3x^2}-x\right). \] Since \[ y=\frac{3x-x^3}{1-3x^2}, \] we get \[ (1-3x^2)\frac{d^2y}{dx^2}-12x\frac{dy}{dx} = 6(y-x). \]
Step 4: Final conclusion.
Therefore, \[ \boxed{6(y-x)}. \]
Given, \[ y=\tan(3\tan^{-1}x). \] Let \[ \theta=\tan^{-1}x. \] Then, \[ x=\tan\theta. \] So, \[ y=\tan 3\theta. \] Using the formula, \[ \tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}, \] we get \[ y=\frac{3x-x^3}{1-3x^2}. \]
Step 2: Differentiate \(y\).
After differentiating, \[ \frac{dy}{dx}=\frac{3(1+x^2)^2}{(1-3x^2)^2}. \] Again differentiating, \[ \frac{d^2y}{dx^2} = \frac{12x(1+x^2)(1-3x^2)+36x(1+x^2)^2}{(1-3x^2)^3}. \]
Step 3: Substitute in the given expression.
Now, \[ (1-3x^2)\frac{d^2y}{dx^2}-12x\frac{dy}{dx} \] simplifies to \[ 6\left(\frac{3x-x^3}{1-3x^2}-x\right). \] Since \[ y=\frac{3x-x^3}{1-3x^2}, \] we get \[ (1-3x^2)\frac{d^2y}{dx^2}-12x\frac{dy}{dx} = 6(y-x). \]
Step 4: Final conclusion.
Therefore, \[ \boxed{6(y-x)}. \]
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