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Answer :
To solve the problem, we need to find the second derivative [tex]\(\frac{d^2 y}{d x^2}\)[/tex] at [tex]\(x = 1\)[/tex] for the function [tex]\(y = f(x)\)[/tex].
First, we know that:
[tex]\[
\frac{d y}{d x} = 4 \sqrt{y^2 + 7x^2}
\][/tex]
We're given the initial condition [tex]\(f(1) = 3\)[/tex]. This means that when [tex]\(x = 1\)[/tex], [tex]\(y = 3\)[/tex].
### Step 1: Calculate [tex]\(\frac{d y}{d x}\)[/tex] at [tex]\(x = 1\)[/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 3\)[/tex] into the expression for [tex]\(\frac{d y}{d x}\)[/tex]:
[tex]\[
\frac{d y}{d x} = 4 \sqrt{3^2 + 7 \cdot 1^2} = 4 \sqrt{9 + 7} = 4 \sqrt{16} = 4 \times 4 = 16
\][/tex]
### Step 2: Differentiate [tex]\(\frac{d y}{d x}\)[/tex] to find [tex]\(\frac{d^2 y}{d x^2}\)[/tex]
Now, we need to differentiate [tex]\(\frac{d y}{d x} = 4 \sqrt{y^2 + 7x^2}\)[/tex] with respect to [tex]\(x\)[/tex].
Using the chain rule, differentiate the right side:
- The derivative of [tex]\(4 \sqrt{y^2 + 7x^2}\)[/tex] using chain rule is:
[tex]\[
\frac{d}{dx} \left( 4 \sqrt{y^2 + 7x^2} \right) = \frac{4}{2\sqrt{y^2 + 7x^2}} \left( 2y \frac{dy}{dx} + 14x \right)
\][/tex]
Simplifying, this becomes:
[tex]\[
\frac{4(y \frac{dy}{dx} + 7x)}{\sqrt{y^2 + 7x^2}}
\][/tex]
### Step 3: Evaluate [tex]\(\frac{d^2 y}{d x^2}\)[/tex] at [tex]\(x = 1\)[/tex]
- We already know [tex]\(\frac{dy}{dx} = 16\)[/tex] at [tex]\(x = 1\)[/tex].
- Substitute [tex]\(y = 3\)[/tex], [tex]\(x = 1\)[/tex], and [tex]\(\frac{dy}{dx} = 16\)[/tex]:
[tex]\[
\frac{4(3 \cdot 16 + 7 \cdot 1)}{\sqrt{3^2 + 7 \cdot 1^2}} = \frac{4(48 + 7)}{\sqrt{16}}
\][/tex]
[tex]\[
= \frac{4 \cdot 55}{4} = 55
\][/tex]
Thus, the value of [tex]\(\frac{d^2 y}{d x^2}\)[/tex] at [tex]\(x = 1\)[/tex] is [tex]\( \boxed{55} \)[/tex].
First, we know that:
[tex]\[
\frac{d y}{d x} = 4 \sqrt{y^2 + 7x^2}
\][/tex]
We're given the initial condition [tex]\(f(1) = 3\)[/tex]. This means that when [tex]\(x = 1\)[/tex], [tex]\(y = 3\)[/tex].
### Step 1: Calculate [tex]\(\frac{d y}{d x}\)[/tex] at [tex]\(x = 1\)[/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 3\)[/tex] into the expression for [tex]\(\frac{d y}{d x}\)[/tex]:
[tex]\[
\frac{d y}{d x} = 4 \sqrt{3^2 + 7 \cdot 1^2} = 4 \sqrt{9 + 7} = 4 \sqrt{16} = 4 \times 4 = 16
\][/tex]
### Step 2: Differentiate [tex]\(\frac{d y}{d x}\)[/tex] to find [tex]\(\frac{d^2 y}{d x^2}\)[/tex]
Now, we need to differentiate [tex]\(\frac{d y}{d x} = 4 \sqrt{y^2 + 7x^2}\)[/tex] with respect to [tex]\(x\)[/tex].
Using the chain rule, differentiate the right side:
- The derivative of [tex]\(4 \sqrt{y^2 + 7x^2}\)[/tex] using chain rule is:
[tex]\[
\frac{d}{dx} \left( 4 \sqrt{y^2 + 7x^2} \right) = \frac{4}{2\sqrt{y^2 + 7x^2}} \left( 2y \frac{dy}{dx} + 14x \right)
\][/tex]
Simplifying, this becomes:
[tex]\[
\frac{4(y \frac{dy}{dx} + 7x)}{\sqrt{y^2 + 7x^2}}
\][/tex]
### Step 3: Evaluate [tex]\(\frac{d^2 y}{d x^2}\)[/tex] at [tex]\(x = 1\)[/tex]
- We already know [tex]\(\frac{dy}{dx} = 16\)[/tex] at [tex]\(x = 1\)[/tex].
- Substitute [tex]\(y = 3\)[/tex], [tex]\(x = 1\)[/tex], and [tex]\(\frac{dy}{dx} = 16\)[/tex]:
[tex]\[
\frac{4(3 \cdot 16 + 7 \cdot 1)}{\sqrt{3^2 + 7 \cdot 1^2}} = \frac{4(48 + 7)}{\sqrt{16}}
\][/tex]
[tex]\[
= \frac{4 \cdot 55}{4} = 55
\][/tex]
Thus, the value of [tex]\(\frac{d^2 y}{d x^2}\)[/tex] at [tex]\(x = 1\)[/tex] is [tex]\( \boxed{55} \)[/tex].
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