# Let y = f(x) be a twice-differentiable function such that f(1) = 2 and . What is the value of (d^2y)/(dx^2) at  x = 1?

Feb 18, 2017

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 28$ when $x = 1$

#### Explanation:

Let $y = f \left(x\right)$ be a twice-differentiable function such that $f \left(1\right) = 2$ and . What is the value of $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$ at$x = 1$?

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} + 3$

We know that $y = 2$ when $x = 1$ and so:

$y = 2 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = {2}^{2} + 3 = 7$

Differentiating the above equation (implicitly) wrt $x$ we get:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$

And so when $x = 1$ we have:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 \left(2\right) \left(7\right) = 28$

Feb 18, 2017

$28$

#### Explanation:

After two derivations, the terms like $a x + b$ disappear. Then if

$\frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} + 3 \to \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 y y ' = 2 y \left({y}^{2} + 3\right)$ but

$y \left(1\right) = 2$ so

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 \times 2 \left({2}^{2} + 3\right) = 28$