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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] of the function [tex]\(y = f(x)\)[/tex] evaluated at [tex]\(x = 1\)[/tex].
Here's a step-by-step explanation of how to achieve this:
1. Understand the Given Derivative:
The problem states that the derivative of [tex]\(y\)[/tex], [tex]\(\frac{d y}{d x}\)[/tex], is given by:
[tex]\[
\frac{d y}{d x} = 4 \sqrt{y^2 + 7x^2}
\][/tex]
2. Substitute the Known Values:
We're given [tex]\(f(1) = 3\)[/tex]. Therefore, when [tex]\(x = 1\)[/tex], [tex]\(y = f(1) = 3\)[/tex].
Substitute [tex]\(y = 3\)[/tex] and [tex]\(x = 1\)[/tex] into the expression for [tex]\(\frac{d y}{d x}\)[/tex]:
[tex]\[
\frac{d y}{d x} = 4 \sqrt{3^2 + 7 \times 1^2} = 4 \sqrt{9 + 7} = 4 \sqrt{16} = 4 \times 4 = 16
\][/tex]
3. Calculate the Second Derivative:
To find [tex]\(\frac{d^2 y}{d x^2}\)[/tex], we differentiate [tex]\(\frac{d y}{d x} = 4 \sqrt{y^2 + 7x^2}\)[/tex] with respect to [tex]\(x\)[/tex].
Using the chain rule, the differentiation of [tex]\(4 \sqrt{y^2 + 7x^2}\)[/tex] gives us:
[tex]\[
\frac{d^2 y}{d x^2} = \frac{d}{d x}\left(4 \sqrt{y^2 + 7x^2}\right)
\][/tex]
To find this, consider that the derivative of [tex]\(\sqrt{y^2 + 7x^2}\)[/tex] with respect to [tex]\(x\)[/tex] involves applying the chain rule:
[tex]\[
= \frac{d}{d x}(y^2 + 7x^2)^{1/2}
\][/tex]
The derivative of the inside function [tex]\(y^2 + 7x^2\)[/tex] with respect to [tex]\(x\)[/tex] gives:
- Differentiate [tex]\(y^2\)[/tex] with respect to [tex]\(x\)[/tex]: [tex]\(\frac{d}{d x}(y^2) = 2y \frac{d y}{d x}\)[/tex]
- Differentiate [tex]\(7x^2\)[/tex] with respect to [tex]\(x\)[/tex]: [tex]\(\frac{d}{d x}(7x^2) = 14x\)[/tex]
Putting it all together:
[tex]\[
\frac{d^2 y}{d x^2} = 4 \times \frac{2y \frac{d y}{d x} + 14x}{2 \sqrt{y^2 + 7x^2}}
\][/tex]
Simplify using substitution at [tex]\(x = 1\)[/tex], [tex]\(y = 3\)[/tex], and [tex]\(\frac{d y}{d x} = 16\)[/tex]:
- Substitute these into the expression:
[tex]\[
\frac{d^2 y}{d x^2} = \frac{8(3)(16) + 56}{2 \times 4}
\][/tex]
[tex]\[
= \frac{384 + 56}{8} = \frac{440}{8} = 55
\][/tex]
Thus, the value of [tex]\(\frac{d^2 y}{d x^2}\)[/tex] at [tex]\(x = 1\)[/tex] is [tex]\(\boxed{55}\)[/tex].
Here's a step-by-step explanation of how to achieve this:
1. Understand the Given Derivative:
The problem states that the derivative of [tex]\(y\)[/tex], [tex]\(\frac{d y}{d x}\)[/tex], is given by:
[tex]\[
\frac{d y}{d x} = 4 \sqrt{y^2 + 7x^2}
\][/tex]
2. Substitute the Known Values:
We're given [tex]\(f(1) = 3\)[/tex]. Therefore, when [tex]\(x = 1\)[/tex], [tex]\(y = f(1) = 3\)[/tex].
Substitute [tex]\(y = 3\)[/tex] and [tex]\(x = 1\)[/tex] into the expression for [tex]\(\frac{d y}{d x}\)[/tex]:
[tex]\[
\frac{d y}{d x} = 4 \sqrt{3^2 + 7 \times 1^2} = 4 \sqrt{9 + 7} = 4 \sqrt{16} = 4 \times 4 = 16
\][/tex]
3. Calculate the Second Derivative:
To find [tex]\(\frac{d^2 y}{d x^2}\)[/tex], we differentiate [tex]\(\frac{d y}{d x} = 4 \sqrt{y^2 + 7x^2}\)[/tex] with respect to [tex]\(x\)[/tex].
Using the chain rule, the differentiation of [tex]\(4 \sqrt{y^2 + 7x^2}\)[/tex] gives us:
[tex]\[
\frac{d^2 y}{d x^2} = \frac{d}{d x}\left(4 \sqrt{y^2 + 7x^2}\right)
\][/tex]
To find this, consider that the derivative of [tex]\(\sqrt{y^2 + 7x^2}\)[/tex] with respect to [tex]\(x\)[/tex] involves applying the chain rule:
[tex]\[
= \frac{d}{d x}(y^2 + 7x^2)^{1/2}
\][/tex]
The derivative of the inside function [tex]\(y^2 + 7x^2\)[/tex] with respect to [tex]\(x\)[/tex] gives:
- Differentiate [tex]\(y^2\)[/tex] with respect to [tex]\(x\)[/tex]: [tex]\(\frac{d}{d x}(y^2) = 2y \frac{d y}{d x}\)[/tex]
- Differentiate [tex]\(7x^2\)[/tex] with respect to [tex]\(x\)[/tex]: [tex]\(\frac{d}{d x}(7x^2) = 14x\)[/tex]
Putting it all together:
[tex]\[
\frac{d^2 y}{d x^2} = 4 \times \frac{2y \frac{d y}{d x} + 14x}{2 \sqrt{y^2 + 7x^2}}
\][/tex]
Simplify using substitution at [tex]\(x = 1\)[/tex], [tex]\(y = 3\)[/tex], and [tex]\(\frac{d y}{d x} = 16\)[/tex]:
- Substitute these into the expression:
[tex]\[
\frac{d^2 y}{d x^2} = \frac{8(3)(16) + 56}{2 \times 4}
\][/tex]
[tex]\[
= \frac{384 + 56}{8} = \frac{440}{8} = 55
\][/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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