Question

Let y=f(x) be a twice-differentiable function such that f(1)=3 and dydx=4y2+7x2−−−−−−−√. What is the value of d2ydx2 at x=1 ?.

Answer 1

Answer:

55

Explanation:

Find the derivative of $\frac{dy}{dx}$ to get $\frac{d^2y}{dx^2}$ .

4$\sqrt{y^2+7x^2}$ = 4($y^{2}$+7$x^{2}$)^1/2

$\frac{d^2y}{dx^2}$ = 4(1/2)($y^{2}$+7$x^{2}$)^-1/2 * (2y$\frac{dy}{dx}$ +14x)

Plug in x=1 and y=3.

2(9+7)^-1/2 * (6$\frac{dy}{dx}$ +14)

2(1/4) * (6$\frac{dy}{dx}$ +14)

(1/2) (6$\frac{dy}{dx}$ +14)

3$\frac{dy}{dx}$ +7

Plug in $\frac{dy}{dx}$.

3(4$\sqrt{y^2+7x^2}$ ) +7

Plug in x=1 and y=3.

3(4$\sqrt{9+7}$ ) +7

3(4$\sqrt{16}$ ) +7

3(4*4)+7

3(16)+7

48+7

55