QuestionLet $y=f(x)$ be a twice-differentiable function such that $f(1)=3$ and $\frac{d y}{d x}=4 \sqrt{y^{2}+7 x^{2}}$. What is the value of $\frac{d^{2} y}{d x^{2}}$ at $x=1$ ? (A) 10 (B) 23 (C) 55 (D) 160

Studdy Solution
Substitute $\frac{d y}{d x} = 16$, $y = 3$, and $x = 1$ into the expression for $\frac{d^{2} y}{d x^{2}}$:
$$\frac{d^{2} y}{d x^{2}} = \frac{4}{\sqrt{3^{2} + 7 \cdot 1^{2}}} \cdot \left( 3 \cdot 16 + 7 \cdot 1 \right)$$
$$\frac{d^{2} y}{d x^{2}} = \frac{4}{\sqrt{16}} \cdot \left( 48 + 7 \right)$$
$$\frac{d^{2} y}{d x^{2}} = \frac{4}{4} \cdot 55$$
$$\frac{d^{2} y}{d x^{2}} = 1 \cdot 55$$
$$\frac{d^{2} y}{d x^{2}} = 55$$
The value of $\frac{d^{2} y}{d x^{2}}$ at $x = 1$ is:
$$\boxed{55}$$