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Answer :
To solve this problem, we are tasked with finding the second derivative of the function [tex]\( y = f(x) \)[/tex] at [tex]\( x = 1 \)[/tex]. We are given the first derivative [tex]\(\frac{dy}{dx} = 4 \sqrt{y^2 + 7x^2} \)[/tex] and the initial condition [tex]\( f(1) = 3 \)[/tex].
Here's the detailed step-by-step explanation:
1. Find the value of [tex]\(\frac{dy}{dx}\)[/tex] at [tex]\( x = 1 \)[/tex]:
Substitute [tex]\( x = 1 \)[/tex] and [tex]\( y = 3 \)[/tex] into the expression for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
\frac{dy}{dx} = 4 \sqrt{3^2 + 7 \cdot 1^2} = 4 \sqrt{9 + 7} = 4 \sqrt{16} = 4 \cdot 4 = 16
\][/tex]
2. Differentiate [tex]\(\frac{dy}{dx}\)[/tex] to find [tex]\(\frac{d^2y}{dx^2}\)[/tex]:
The expression for [tex]\(\frac{dy}{dx}\)[/tex] is [tex]\( 4 \sqrt{y^2 + 7x^2} \)[/tex]. To find the second derivative, we need to use the chain rule and implicit differentiation.
Let [tex]\( u = y^2 + 7x^2 \)[/tex]. So, [tex]\(\frac{dy}{dx} = 4\sqrt{u}\)[/tex].
Differentiate [tex]\( \frac{dy}{dx} \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
\frac{d}{dx}(4\sqrt{u}) = 4 \cdot \frac{d}{dx}(u^{1/2})
\][/tex]
Using the chain rule, [tex]\(\frac{d}{dx}(u^{1/2}) = \frac{1}{2}u^{-1/2} \cdot \frac{du}{dx}\)[/tex].
Therefore:
[tex]\[
\frac{d^2y}{dx^2} = 4 \cdot \frac{1}{2} \cdot u^{-1/2} \cdot \frac{du}{dx} = 2 \cdot \frac{du}{dx} \cdot \frac{1}{\sqrt{u}}
\][/tex]
3. Find [tex]\(\frac{du}{dx}\)[/tex] at [tex]\(x = 1\)[/tex]:
Differentiate [tex]\( u = y^2 + 7x^2 \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
\frac{du}{dx} = 2y \cdot \frac{dy}{dx} + 14x
\][/tex]
Substitute the known values [tex]\( y = 3 \)[/tex], [tex]\( \frac{dy}{dx} = 16 \)[/tex], and [tex]\( x = 1 \)[/tex]:
[tex]\[
\frac{du}{dx} = 2 \cdot 3 \cdot 16 + 14 \cdot 1 = 96 + 14 = 110
\][/tex]
4. Calculate [tex]\(\frac{d^2y}{dx^2}\)[/tex] at [tex]\(x = 1\)[/tex]:
Substitute back into the expression for [tex]\(\frac{d^2y}{dx^2}\)[/tex]:
[tex]\[
\frac{d^2y}{dx^2} = 2 \cdot \frac{110}{\sqrt{16}} = 2 \cdot \frac{110}{4} = 2 \cdot 27.5 = 55
\][/tex]
Thus, the value of [tex]\(\frac{d^2y}{dx^2}\)[/tex] at [tex]\( x = 1 \)[/tex] is [tex]\( \boxed{55} \)[/tex].
Here's the detailed step-by-step explanation:
1. Find the value of [tex]\(\frac{dy}{dx}\)[/tex] at [tex]\( x = 1 \)[/tex]:
Substitute [tex]\( x = 1 \)[/tex] and [tex]\( y = 3 \)[/tex] into the expression for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
\frac{dy}{dx} = 4 \sqrt{3^2 + 7 \cdot 1^2} = 4 \sqrt{9 + 7} = 4 \sqrt{16} = 4 \cdot 4 = 16
\][/tex]
2. Differentiate [tex]\(\frac{dy}{dx}\)[/tex] to find [tex]\(\frac{d^2y}{dx^2}\)[/tex]:
The expression for [tex]\(\frac{dy}{dx}\)[/tex] is [tex]\( 4 \sqrt{y^2 + 7x^2} \)[/tex]. To find the second derivative, we need to use the chain rule and implicit differentiation.
Let [tex]\( u = y^2 + 7x^2 \)[/tex]. So, [tex]\(\frac{dy}{dx} = 4\sqrt{u}\)[/tex].
Differentiate [tex]\( \frac{dy}{dx} \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
\frac{d}{dx}(4\sqrt{u}) = 4 \cdot \frac{d}{dx}(u^{1/2})
\][/tex]
Using the chain rule, [tex]\(\frac{d}{dx}(u^{1/2}) = \frac{1}{2}u^{-1/2} \cdot \frac{du}{dx}\)[/tex].
Therefore:
[tex]\[
\frac{d^2y}{dx^2} = 4 \cdot \frac{1}{2} \cdot u^{-1/2} \cdot \frac{du}{dx} = 2 \cdot \frac{du}{dx} \cdot \frac{1}{\sqrt{u}}
\][/tex]
3. Find [tex]\(\frac{du}{dx}\)[/tex] at [tex]\(x = 1\)[/tex]:
Differentiate [tex]\( u = y^2 + 7x^2 \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
\frac{du}{dx} = 2y \cdot \frac{dy}{dx} + 14x
\][/tex]
Substitute the known values [tex]\( y = 3 \)[/tex], [tex]\( \frac{dy}{dx} = 16 \)[/tex], and [tex]\( x = 1 \)[/tex]:
[tex]\[
\frac{du}{dx} = 2 \cdot 3 \cdot 16 + 14 \cdot 1 = 96 + 14 = 110
\][/tex]
4. Calculate [tex]\(\frac{d^2y}{dx^2}\)[/tex] at [tex]\(x = 1\)[/tex]:
Substitute back into the expression for [tex]\(\frac{d^2y}{dx^2}\)[/tex]:
[tex]\[
\frac{d^2y}{dx^2} = 2 \cdot \frac{110}{\sqrt{16}} = 2 \cdot \frac{110}{4} = 2 \cdot 27.5 = 55
\][/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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