Question

If $y = (\tan^{-1}x)^2$ then $(x^2 + 1)^2\frac{d^2y}{dx^2} + 2x(x^2 + 1)\frac{dy}{dx} =$

• A. 4
• B. 2
• C. 1
• D. 0

Answer 1

Answer: (B)

2

Answer 2

The problem asks us to find the value of the expression $(x^2 + 1)^2\frac{d^2y}{dx^2} + 2x(x^2 + 1)\frac{dy}{dx}$ given that $y = (\tan^{-1}x)^2$.

Step 1: Find the first derivative $\frac{dy}{dx}$. Given $y = (\tan^{-1}x)^2$. Using the chain rule, $\frac{dy}{dx} = 2(\tan^{-1}x) \cdot \frac{d}{dx}(\tan^{-1}x)$. We know that $\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}$. So, $\frac{dy}{dx} = 2(\tan^{-1}x) \cdot \frac{1}{1+x^2}$ Rearrange this equation by multiplying both sides by $(1+x^2)$: $(1+x^2)\frac{dy}{dx} = 2\tan^{-1}x \quad \text{(Equation 1)}$

Step 2: Find the second derivative $\frac{d^2y}{dx^2}$. Differentiate both sides of Equation 1 with respect to $x$. For the left side, use the product rule $\frac{d}{dx}(uv) = u'v + uv'$ where $u = (1+x^2)$ and $v = \frac{dy}{dx}$. $\frac{d}{dx}\left((1+x^2)\frac{dy}{dx}\right) = \frac{d}{dx}(1+x^2) \cdot \frac{dy}{dx} + (1+x^2) \cdot \frac{d}{dx}\left(\frac{dy}{dx}\right)$ $= 2x \frac{dy}{dx} + (1+x^2) \frac{d^2y}{dx^2}$ For the right side, differentiate $2\tan^{-1}x$: $\frac{d}{dx}(2\tan^{-1}x) = 2 \cdot \frac{1}{1+x^2}$ Equating the derivatives of both sides: $2x \frac{dy}{dx} + (1+x^2) \frac{d^2y}{dx^2} = \frac{2}{1+x^2} \quad \text{(Equation 2)}$

Step 3: Obtain the required expression. The expression we need to find is $(x^2 + 1)^2\frac{d^2y}{dx^2} + 2x(x^2 + 1)\frac{dy}{dx}$. Notice that if we multiply Equation 2 by $(1+x^2)$, we will get the required expression. Multiply both sides of Equation 2 by $(1+x^2)$: $(1+x^2) \left[ 2x \frac{dy}{dx} + (1+x^2) \frac{d^2y}{dx^2} \right] = (1+x^2) \left[ \frac{2}{1+x^2} \right]$ Distribute $(1+x^2)$ on the left side: $2x(1+x^2)\frac{dy}{dx} + (1+x^2)^2\frac{d^2y}{dx^2} = 2$ Rearranging the terms to match the question's format: $(x^2 + 1)^2\frac{d^2y}{dx^2} + 2x(x^2 + 1)\frac{dy}{dx} = 2$

The value of the expression is 2.