Question

If $\mathbf{x =}\int_{\mathbf{0}}^{\mathbf{y}}\frac{\mathbf{1}}{\sqrt{\mathbf{1 + 4}\mathbf{t}^{\mathbf{2}}}}\mathbf{dt}$, then $\frac{d^{2}y}{dx^{2}}$ is

• A. 2y
• B. 4y
• C. 8y
• D. 6 y

Answer 1

Answer: (B)

4y

Answer 2

$x = \int_{0}^{y}{\frac{1}{\sqrt{1 + 4t^{2}}}dt}$ ⇒ $\frac{dx}{dy} = \frac{1}{\sqrt{1 + 4y^{2}}}$ ⇒ $\frac{dy}{dx} = \sqrt{1 + 4y^{2}}$

⇒ $\frac{d^{2}y}{dx^{2}} = \frac{4y}{\sqrt{1 + 4y^{2}}}\frac{dy}{dx}$ ⇒ $\frac{d^{2}y}{dx^{2}} = \frac{4y}{\sqrt{1 + 4y^{2}}} \cdot \sqrt{1 + 4y^{2}} = 4y$