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Answer :
We are given a twice-differentiable function [tex]$y=f(x)$[/tex] with
[tex]$$
f(1)=3
$$[/tex]
and
[tex]$$
\frac{dy}{dx}=4\sqrt{y^2+7x^2}.
$$[/tex]
We want to find the value of the second derivative, [tex]$\frac{d^2y}{dx^2}$[/tex], at [tex]$x=1$[/tex].
1. Define
[tex]$$
u = y^2 + 7x^2.
$$[/tex]
Then the first derivative is written as
[tex]$$
\frac{dy}{dx}=4\sqrt{u}.
$$[/tex]
2. Differentiate the first derivative with respect to [tex]$x$[/tex]. Using the chain rule,
[tex]$$
\frac{d^2y}{dx^2} = \frac{d}{dx}\left( 4 \sqrt{u} \right)
= 4\cdot\frac{d}{dx}\left(\sqrt{u}\right)
= 4 \cdot \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}
= \frac{2}{\sqrt{u}}\,\frac{du}{dx}.
$$[/tex]
3. Now, compute [tex]$\frac{du}{dx}$[/tex]. Since
[tex]$$
u = y^2 + 7x^2,
$$[/tex]
differentiate both terms with respect to [tex]$x$[/tex]:
[tex]$$
\frac{du}{dx} = \frac{d}{dx}(y^2) + \frac{d}{dx}(7x^2)
= 2y\,\frac{dy}{dx} + 14x.
$$[/tex]
4. Substitute [tex]$\frac{dy}{dx}=4\sqrt{u}$[/tex] into the above expression:
[tex]$$
\frac{du}{dx} = 2y\,(4\sqrt{u}) + 14x
= 8y\sqrt{u} + 14x.
$$[/tex]
5. Replace [tex]$\frac{du}{dx}$[/tex] back into the expression for [tex]$\frac{d^2y}{dx^2}$[/tex]:
[tex]$$
\frac{d^2y}{dx^2} = \frac{2}{\sqrt{u}}\,(8y\sqrt{u} + 14x).
$$[/tex]
Notice that [tex]$\sqrt{u}$[/tex] cancels with [tex]$\sqrt{u}$[/tex] in the term [tex]$8y\sqrt{u}$[/tex]:
[tex]$$
\frac{d^2y}{dx^2} = \frac{16y\sqrt{u} + 28x}{\sqrt{u}}.
$$[/tex]
6. Now, evaluate at [tex]$x=1$[/tex] and [tex]$y=f(1)=3$[/tex]. First, compute [tex]$u$[/tex]:
[tex]$$
u = y^2 + 7x^2 = 3^2 + 7\cdot1^2 = 9 + 7 = 16.
$$[/tex]
Thus, [tex]$\sqrt{u}=4$[/tex].
7. Substitute [tex]$x=1$[/tex], [tex]$y=3$[/tex], and [tex]$\sqrt{u}=4$[/tex] into the expression for [tex]$\frac{d^2y}{dx^2}$[/tex]:
[tex]$$
\frac{d^2y}{dx^2} = \frac{16 \cdot 3 \cdot 4 + 28 \cdot 1}{4}.
$$[/tex]
Simplify step-by-step:
[tex]$$
16 \cdot 3 \cdot 4 = 192,
$$[/tex]
[tex]$$
\frac{192 + 28}{4} = \frac{220}{4} = 55.
$$[/tex]
Thus, the value of [tex]$\frac{d^2y}{dx^2}$[/tex] at [tex]$x=1$[/tex] is
[tex]$$
\boxed{55}.
$$[/tex]
[tex]$$
f(1)=3
$$[/tex]
and
[tex]$$
\frac{dy}{dx}=4\sqrt{y^2+7x^2}.
$$[/tex]
We want to find the value of the second derivative, [tex]$\frac{d^2y}{dx^2}$[/tex], at [tex]$x=1$[/tex].
1. Define
[tex]$$
u = y^2 + 7x^2.
$$[/tex]
Then the first derivative is written as
[tex]$$
\frac{dy}{dx}=4\sqrt{u}.
$$[/tex]
2. Differentiate the first derivative with respect to [tex]$x$[/tex]. Using the chain rule,
[tex]$$
\frac{d^2y}{dx^2} = \frac{d}{dx}\left( 4 \sqrt{u} \right)
= 4\cdot\frac{d}{dx}\left(\sqrt{u}\right)
= 4 \cdot \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}
= \frac{2}{\sqrt{u}}\,\frac{du}{dx}.
$$[/tex]
3. Now, compute [tex]$\frac{du}{dx}$[/tex]. Since
[tex]$$
u = y^2 + 7x^2,
$$[/tex]
differentiate both terms with respect to [tex]$x$[/tex]:
[tex]$$
\frac{du}{dx} = \frac{d}{dx}(y^2) + \frac{d}{dx}(7x^2)
= 2y\,\frac{dy}{dx} + 14x.
$$[/tex]
4. Substitute [tex]$\frac{dy}{dx}=4\sqrt{u}$[/tex] into the above expression:
[tex]$$
\frac{du}{dx} = 2y\,(4\sqrt{u}) + 14x
= 8y\sqrt{u} + 14x.
$$[/tex]
5. Replace [tex]$\frac{du}{dx}$[/tex] back into the expression for [tex]$\frac{d^2y}{dx^2}$[/tex]:
[tex]$$
\frac{d^2y}{dx^2} = \frac{2}{\sqrt{u}}\,(8y\sqrt{u} + 14x).
$$[/tex]
Notice that [tex]$\sqrt{u}$[/tex] cancels with [tex]$\sqrt{u}$[/tex] in the term [tex]$8y\sqrt{u}$[/tex]:
[tex]$$
\frac{d^2y}{dx^2} = \frac{16y\sqrt{u} + 28x}{\sqrt{u}}.
$$[/tex]
6. Now, evaluate at [tex]$x=1$[/tex] and [tex]$y=f(1)=3$[/tex]. First, compute [tex]$u$[/tex]:
[tex]$$
u = y^2 + 7x^2 = 3^2 + 7\cdot1^2 = 9 + 7 = 16.
$$[/tex]
Thus, [tex]$\sqrt{u}=4$[/tex].
7. Substitute [tex]$x=1$[/tex], [tex]$y=3$[/tex], and [tex]$\sqrt{u}=4$[/tex] into the expression for [tex]$\frac{d^2y}{dx^2}$[/tex]:
[tex]$$
\frac{d^2y}{dx^2} = \frac{16 \cdot 3 \cdot 4 + 28 \cdot 1}{4}.
$$[/tex]
Simplify step-by-step:
[tex]$$
16 \cdot 3 \cdot 4 = 192,
$$[/tex]
[tex]$$
\frac{192 + 28}{4} = \frac{220}{4} = 55.
$$[/tex]
Thus, the value of [tex]$\frac{d^2y}{dx^2}$[/tex] at [tex]$x=1$[/tex] is
[tex]$$
\boxed{55}.
$$[/tex]
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