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Answer :
To solve this problem, we need to find the second derivative [tex]\(\frac{d^2 y}{d x^2}\)[/tex] at [tex]\(x=1\)[/tex] given the differential equation [tex]\(\frac{dy}{dx} = 4 \sqrt{y^2 + 7x^2}\)[/tex] and the condition [tex]\(f(1) = 3\)[/tex].
Step 1: Differentiate the given equation
Starting with [tex]\(\frac{dy}{dx} = 4 \sqrt{y^2 + 7x^2}\)[/tex], we differentiate both sides with respect to [tex]\(x\)[/tex]:
[tex]\[
\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(4 \sqrt{y^2 + 7x^2}\right)
\][/tex]
The left side [tex]\(\frac{d}{dx}\left(\frac{dy}{dx}\right)\)[/tex] is the second derivative, [tex]\(\frac{d^2y}{dx^2}\)[/tex].
Step 2: Differentiate the right side using the chain rule
Using the chain rule on the right side:
[tex]\[
\frac{d}{dx}\left(4 \sqrt{y^2 + 7x^2}\right) = 4 \cdot \frac{1}{2\sqrt{y^2 + 7x^2}} \cdot \left(2y \cdot \frac{dy}{dx} + 14x\right)
\][/tex]
Simplifying gives us:
[tex]\[
= \frac{4(y \cdot \frac{dy}{dx} + 7x)}{\sqrt{y^2 + 7x^2}}
\][/tex]
So,
[tex]\[
\frac{d^2 y}{d x^2} = \frac{4(y \cdot \frac{dy}{dx} + 7x)}{\sqrt{y^2 + 7x^2}}
\][/tex]
Step 3: Evaluate at [tex]\(x=1\)[/tex] and [tex]\(y=3\)[/tex]
We know that when [tex]\(x = 1\)[/tex], [tex]\(y = 3\)[/tex].
First, substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 3\)[/tex] into the expression for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
\frac{dy}{dx}\bigg|_{x=1, y=3} = 4 \sqrt{3^2 + 7 \cdot 1^2} = 4 \sqrt{9 + 7} = 4 \sqrt{16} = 16
\][/tex]
Now substitute into the expression for [tex]\(\frac{d^2 y}{dx^2}\)[/tex]:
[tex]\[
\frac{d^2 y}{d x^2}\bigg|_{x=1, y=3, \frac{dy}{dx}=16} = \frac{4(3 \cdot 16 + 7 \cdot 1)}{\sqrt{9 + 7}}
\][/tex]
Calculate the numerator:
[tex]\[
3 \cdot 16 = 48, \quad 7 \cdot 1 = 7, \quad 48 + 7 = 55.
\][/tex]
Substitute and simplify:
[tex]\[
= \frac{4 \cdot 55}{\sqrt{16}} = \frac{220}{4} = 55.
\][/tex]
Therefore, [tex]\(\frac{d^2 y}{d x^2}\)[/tex] at [tex]\(x=1\)[/tex] is [tex]\(55\)[/tex].
The correct answer is (C) 55.
Step 1: Differentiate the given equation
Starting with [tex]\(\frac{dy}{dx} = 4 \sqrt{y^2 + 7x^2}\)[/tex], we differentiate both sides with respect to [tex]\(x\)[/tex]:
[tex]\[
\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(4 \sqrt{y^2 + 7x^2}\right)
\][/tex]
The left side [tex]\(\frac{d}{dx}\left(\frac{dy}{dx}\right)\)[/tex] is the second derivative, [tex]\(\frac{d^2y}{dx^2}\)[/tex].
Step 2: Differentiate the right side using the chain rule
Using the chain rule on the right side:
[tex]\[
\frac{d}{dx}\left(4 \sqrt{y^2 + 7x^2}\right) = 4 \cdot \frac{1}{2\sqrt{y^2 + 7x^2}} \cdot \left(2y \cdot \frac{dy}{dx} + 14x\right)
\][/tex]
Simplifying gives us:
[tex]\[
= \frac{4(y \cdot \frac{dy}{dx} + 7x)}{\sqrt{y^2 + 7x^2}}
\][/tex]
So,
[tex]\[
\frac{d^2 y}{d x^2} = \frac{4(y \cdot \frac{dy}{dx} + 7x)}{\sqrt{y^2 + 7x^2}}
\][/tex]
Step 3: Evaluate at [tex]\(x=1\)[/tex] and [tex]\(y=3\)[/tex]
We know that when [tex]\(x = 1\)[/tex], [tex]\(y = 3\)[/tex].
First, substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 3\)[/tex] into the expression for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
\frac{dy}{dx}\bigg|_{x=1, y=3} = 4 \sqrt{3^2 + 7 \cdot 1^2} = 4 \sqrt{9 + 7} = 4 \sqrt{16} = 16
\][/tex]
Now substitute into the expression for [tex]\(\frac{d^2 y}{dx^2}\)[/tex]:
[tex]\[
\frac{d^2 y}{d x^2}\bigg|_{x=1, y=3, \frac{dy}{dx}=16} = \frac{4(3 \cdot 16 + 7 \cdot 1)}{\sqrt{9 + 7}}
\][/tex]
Calculate the numerator:
[tex]\[
3 \cdot 16 = 48, \quad 7 \cdot 1 = 7, \quad 48 + 7 = 55.
\][/tex]
Substitute and simplify:
[tex]\[
= \frac{4 \cdot 55}{\sqrt{16}} = \frac{220}{4} = 55.
\][/tex]
Therefore, [tex]\(\frac{d^2 y}{d x^2}\)[/tex] at [tex]\(x=1\)[/tex] is [tex]\(55\)[/tex].
The correct answer is (C) 55.
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