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Answer :
To solve this problem, we need to find the second derivative [tex]\(\frac{d^2 y}{dx^2}\)[/tex] at [tex]\(x = 1\)[/tex]. We are given that [tex]\(f(1) = 3\)[/tex] and [tex]\(\frac{dy}{dx} = 4\sqrt{y^2 + 7x^2}\)[/tex].
Step 1: Differentiate [tex]\(\frac{dy}{dx}\)[/tex] with respect to [tex]\(x\)[/tex]
To find the second derivative, let's start by differentiating the expression for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
\frac{dy}{dx} = 4\sqrt{y^2 + 7x^2}
\][/tex]
We need to differentiate this with respect to [tex]\(x\)[/tex]. Using the chain rule for differentiation:
[tex]\[
\frac{d^2 y}{dx^2} = 4 \cdot \frac{d}{dx} \left( \sqrt{y^2 + 7x^2} \right)
\][/tex]
The derivative of [tex]\(\sqrt{y^2 + 7x^2}\)[/tex] is:
[tex]\[
\frac{1}{2\sqrt{y^2 + 7x^2}} \cdot \left( 2y \cdot \frac{dy}{dx} + 14x \right)
\][/tex]
Therefore, the expression for [tex]\(\frac{d^2 y}{dx^2}\)[/tex] becomes:
[tex]\[
\frac{d^2 y}{dx^2} = 4 \cdot \frac{1}{2\sqrt{y^2 + 7x^2}} \left( 2y \cdot \frac{dy}{dx} + 14x \right)
\][/tex]
This simplifies to:
[tex]\[
\frac{d^2 y}{dx^2} = \frac{2(2y \cdot \frac{dy}{dx} + 14x)}{\sqrt{y^2 + 7x^2}}
\][/tex]
Step 2: Evaluate [tex]\(\frac{d^2 y}{dx^2}\)[/tex] at [tex]\(x = 1\)[/tex]
We need to substitute [tex]\(x = 1\)[/tex] and use [tex]\(f(1) = y=3\)[/tex]. Also, from [tex]\(\frac{dy}{dx} = 4 \sqrt{y^2 + 7x^2}\)[/tex], when [tex]\(x = 1\)[/tex] and [tex]\(y = 3\)[/tex], we have:
[tex]\[
\frac{dy}{dx} = 4 \sqrt{3^2 + 7 \times 1^2} = 4 \sqrt{9 + 7} = 4 \sqrt{16} = 16
\][/tex]
Substituting these values back into the second derivative expression:
[tex]\[
\frac{d^2 y}{dx^2} = \frac{2\left(2 \times 3 \times 16 + 14 \times 1\right)}{\sqrt{3^2 + 7 \times 1^2}}
\][/tex]
Calculating terms:
- [tex]\(2 \times 3 \times 16 = 96\)[/tex]
- [tex]\(14 \times 1 = 14\)[/tex]
So,
[tex]\[
2 \left( 96 + 14 \right) = 2 \times 110 = 220
\][/tex]
And since the denominator simplifies to [tex]\(\sqrt{16} = 4\)[/tex],
[tex]\[
\frac{d^2 y}{dx^2} = \frac{220}{4} = 55
\][/tex]
Therefore, the value of [tex]\(\frac{d^2 y}{dx^2}\)[/tex] at [tex]\(x = 1\)[/tex] is [tex]\(\boxed{55}\)[/tex].
Step 1: Differentiate [tex]\(\frac{dy}{dx}\)[/tex] with respect to [tex]\(x\)[/tex]
To find the second derivative, let's start by differentiating the expression for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
\frac{dy}{dx} = 4\sqrt{y^2 + 7x^2}
\][/tex]
We need to differentiate this with respect to [tex]\(x\)[/tex]. Using the chain rule for differentiation:
[tex]\[
\frac{d^2 y}{dx^2} = 4 \cdot \frac{d}{dx} \left( \sqrt{y^2 + 7x^2} \right)
\][/tex]
The derivative of [tex]\(\sqrt{y^2 + 7x^2}\)[/tex] is:
[tex]\[
\frac{1}{2\sqrt{y^2 + 7x^2}} \cdot \left( 2y \cdot \frac{dy}{dx} + 14x \right)
\][/tex]
Therefore, the expression for [tex]\(\frac{d^2 y}{dx^2}\)[/tex] becomes:
[tex]\[
\frac{d^2 y}{dx^2} = 4 \cdot \frac{1}{2\sqrt{y^2 + 7x^2}} \left( 2y \cdot \frac{dy}{dx} + 14x \right)
\][/tex]
This simplifies to:
[tex]\[
\frac{d^2 y}{dx^2} = \frac{2(2y \cdot \frac{dy}{dx} + 14x)}{\sqrt{y^2 + 7x^2}}
\][/tex]
Step 2: Evaluate [tex]\(\frac{d^2 y}{dx^2}\)[/tex] at [tex]\(x = 1\)[/tex]
We need to substitute [tex]\(x = 1\)[/tex] and use [tex]\(f(1) = y=3\)[/tex]. Also, from [tex]\(\frac{dy}{dx} = 4 \sqrt{y^2 + 7x^2}\)[/tex], when [tex]\(x = 1\)[/tex] and [tex]\(y = 3\)[/tex], we have:
[tex]\[
\frac{dy}{dx} = 4 \sqrt{3^2 + 7 \times 1^2} = 4 \sqrt{9 + 7} = 4 \sqrt{16} = 16
\][/tex]
Substituting these values back into the second derivative expression:
[tex]\[
\frac{d^2 y}{dx^2} = \frac{2\left(2 \times 3 \times 16 + 14 \times 1\right)}{\sqrt{3^2 + 7 \times 1^2}}
\][/tex]
Calculating terms:
- [tex]\(2 \times 3 \times 16 = 96\)[/tex]
- [tex]\(14 \times 1 = 14\)[/tex]
So,
[tex]\[
2 \left( 96 + 14 \right) = 2 \times 110 = 220
\][/tex]
And since the denominator simplifies to [tex]\(\sqrt{16} = 4\)[/tex],
[tex]\[
\frac{d^2 y}{dx^2} = \frac{220}{4} = 55
\][/tex]
Therefore, the value of [tex]\(\frac{d^2 y}{dx^2}\)[/tex] at [tex]\(x = 1\)[/tex] is [tex]\(\boxed{55}\)[/tex].
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