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Answer :
To solve the problem of finding 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], with the given information:
1. [tex]\(f(1) = 3\)[/tex].
2. [tex]\(\frac{d y}{d x} = 4 \sqrt{y^2 + 7x^2}\)[/tex].
Let's begin to find the expression for the second derivative.
### Step 1: Differentiate the First Derivative
Given:
[tex]\[
\frac{dy}{dx} = 4 \sqrt{y^2 + 7x^2}
\][/tex]
To find the second derivative, we differentiate [tex]\(\frac{dy}{dx}\)[/tex] with respect to [tex]\(x\)[/tex]. Apply the chain rule and the product rule as needed.
### Step 2: Use Chain Rule
Let's denote:
- [tex]\(u = y^2 + 7x^2\)[/tex]
- That makes: [tex]\(\frac{dy}{dx} = 4\sqrt{u}\)[/tex]
Now, differentiate [tex]\(\frac{dy}{dx}\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[
\frac{d^2y}{dx^2} = \frac{d}{dx} \left( 4\sqrt{u} \right)
\][/tex]
Apply the chain rule:
[tex]\[
\frac{d}{dx}(4\sqrt{u}) = 4 \cdot \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} = \frac{2}{\sqrt{u}} \cdot \frac{du}{dx}
\][/tex]
Now find [tex]\(\frac{du}{dx}\)[/tex]:
[tex]\[
\frac{du}{dx} = \frac{d}{dx}(y^2 + 7x^2)
= \frac{d}{dx}(y^2) + \frac{d}{dx}(7x^2)
\][/tex]
By the chain rule and power rule:
- [tex]\(\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}\)[/tex]
- [tex]\(\frac{d}{dx}(7x^2) = 14x\)[/tex]
Thus:
[tex]\[
\frac{du}{dx} = 2y \cdot \frac{dy}{dx} + 14x
\][/tex]
Thus our expression for [tex]\(\frac{d^2y}{dx^2}\)[/tex] becomes:
[tex]\[
\frac{d^2y}{dx^2} = \frac{2}{\sqrt{y^2 + 7x^2}} \cdot (2y \cdot \frac{dy}{dx} + 14x)
\][/tex]
### Step 3: Evaluate at [tex]\(x = 1\)[/tex] and [tex]\(y = 3\)[/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 3\)[/tex] into the expression:
[tex]\[
u = y^2 + 7x^2 = 3^2 + 7(1)^2 = 9 + 7 = 16
\][/tex]
[tex]\(
\frac{dy}{dx} \text{ at } (x, y) = (1, 3) \)[/tex] can also be found from:
[tex]\[
\frac{dy}{dx} = 4 \sqrt{9+7} = 4 \cdot 4 = 16
\][/tex]
Substitute the values into the formula:
[tex]\[
\frac{d^2y}{dx^2} = \frac{2}{4} \cdot (2 \times 3 \times 16 + 14 \times 1) = \frac{1}{2} \cdot (96 + 14) = \frac{1}{2} \cdot 110 = 55
\][/tex]
Therefore, the value of [tex]\(\frac{d^2y}{dx^2}\)[/tex] at [tex]\(x = 1\)[/tex] is [tex]\(\boxed{55}\)[/tex].
1. [tex]\(f(1) = 3\)[/tex].
2. [tex]\(\frac{d y}{d x} = 4 \sqrt{y^2 + 7x^2}\)[/tex].
Let's begin to find the expression for the second derivative.
### Step 1: Differentiate the First Derivative
Given:
[tex]\[
\frac{dy}{dx} = 4 \sqrt{y^2 + 7x^2}
\][/tex]
To find the second derivative, we differentiate [tex]\(\frac{dy}{dx}\)[/tex] with respect to [tex]\(x\)[/tex]. Apply the chain rule and the product rule as needed.
### Step 2: Use Chain Rule
Let's denote:
- [tex]\(u = y^2 + 7x^2\)[/tex]
- That makes: [tex]\(\frac{dy}{dx} = 4\sqrt{u}\)[/tex]
Now, differentiate [tex]\(\frac{dy}{dx}\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[
\frac{d^2y}{dx^2} = \frac{d}{dx} \left( 4\sqrt{u} \right)
\][/tex]
Apply the chain rule:
[tex]\[
\frac{d}{dx}(4\sqrt{u}) = 4 \cdot \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} = \frac{2}{\sqrt{u}} \cdot \frac{du}{dx}
\][/tex]
Now find [tex]\(\frac{du}{dx}\)[/tex]:
[tex]\[
\frac{du}{dx} = \frac{d}{dx}(y^2 + 7x^2)
= \frac{d}{dx}(y^2) + \frac{d}{dx}(7x^2)
\][/tex]
By the chain rule and power rule:
- [tex]\(\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}\)[/tex]
- [tex]\(\frac{d}{dx}(7x^2) = 14x\)[/tex]
Thus:
[tex]\[
\frac{du}{dx} = 2y \cdot \frac{dy}{dx} + 14x
\][/tex]
Thus our expression for [tex]\(\frac{d^2y}{dx^2}\)[/tex] becomes:
[tex]\[
\frac{d^2y}{dx^2} = \frac{2}{\sqrt{y^2 + 7x^2}} \cdot (2y \cdot \frac{dy}{dx} + 14x)
\][/tex]
### Step 3: Evaluate at [tex]\(x = 1\)[/tex] and [tex]\(y = 3\)[/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 3\)[/tex] into the expression:
[tex]\[
u = y^2 + 7x^2 = 3^2 + 7(1)^2 = 9 + 7 = 16
\][/tex]
[tex]\(
\frac{dy}{dx} \text{ at } (x, y) = (1, 3) \)[/tex] can also be found from:
[tex]\[
\frac{dy}{dx} = 4 \sqrt{9+7} = 4 \cdot 4 = 16
\][/tex]
Substitute the values into the formula:
[tex]\[
\frac{d^2y}{dx^2} = \frac{2}{4} \cdot (2 \times 3 \times 16 + 14 \times 1) = \frac{1}{2} \cdot (96 + 14) = \frac{1}{2} \cdot 110 = 55
\][/tex]
Therefore, the value of [tex]\(\frac{d^2y}{dx^2}\)[/tex] at [tex]\(x = 1\)[/tex] is [tex]\(\boxed{55}\)[/tex].
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