Question

If $y = \tan^{-1} \left[ \frac{12x - 64x^3}{1 - 48x^2} \right]$, then $dy/dx = \dots$

• A. $3/(1 + 16x^2)$
• B. $4/(1 + 16x^2)$
• C. $12/(1 + 16x^2)$
• D. $1/(1 + 16x^2)$

Hint:

Whenever you see a $1$ minus "something squared" in the denominator inside an inverse tangent, immediately look for the $\tan(2\theta)$ or $\tan(3\theta)$ substitution patterns. It turns a nightmare chain-rule problem into a 10-second derivative.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We use trigonometric substitution. The expression looks like the formula for $\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$.

Step 2: Formula Application:
Let $4x = \tan \theta$. Then $y = \tan^{-1} \left[ \frac{3(4x) - (4x)^3}{1 - 3(4x)^2} \right] = \tan^{-1} \left[ \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} \right]$.

Step 3: Explanation:
$y = \tan^{-1}(\tan 3\theta) = 3\theta$. Since $4x = \tan \theta \implies \theta = \tan^{-1}(4x)$. $y = 3 \tan^{-1}(4x)$. Differentiating w.r.t $x$: $\frac{dy}{dx} = 3 \cdot \frac{1}{1 + (4x)^2} \cdot 4 = \frac{12}{1 + 16x^2}$.

Step 4: Final Answer:
The derivative is $12/(1 + 16x^2)$.