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

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

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Answer 1 · 2017-02-18T22:57:40

$\frac{d^{2} y}{{d x}^{2}} = 28$ $(d^2y)/(dx^2) = 28$ when $x = 1$ $x=1$

Explanation:

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

We have:

$\frac{d y}{d x} = y^{2} + 3$ $dy/dx = y^2+3$

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

$y = 2 \Rightarrow \frac{d y}{d x} = 2^{2} + 3 = 7$ $y=2 => dy/dx = 2^2+3=7$

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

$\frac{d^{2} y}{{d x}^{2}} = 2 y \frac{d y}{d x}$ $(d^2y)/(dx^2) = 2ydy/dx$

And so when $x = 1$ $x=1$ we have:

$\frac{d^{2} y}{{d x}^{2}} = 2 (2) (7) = 28$ $(d^2y)/(dx^2) = 2(2)(7) = 28$

Answer 2 · 2017-02-18T23:09:16

$28$ $28$

Explanation:

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

$(dy)/(dx)=y^2+3->(d^2y)/(dx^2)=2y y'=2y(y^2+3)$ but

$y(1)=2$ so

$(d^2y)/(dx^2)=2 xx 2(2^2+3)=28$

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

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Let y = f(x)y=f(x)y = f(x) be a twice-differentiable function such that f(1) = 2f(1)=2f(1) = 2 and . What is the value of (d^2y)/(dx^2)d2ydx2(d^2y)/(dx^2) at x = 1x=1 x = 1?

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