Hello again! This is another Calculus question to be explained.

The prompt reads that "If f(x) is a twice-differentiable function such that f(2) = 2 and [tex]\frac{dy}{dx}[/tex] = [tex]6\sqrt{x^2 + 3y^2}[/tex], then what is the value of [tex]\frac{d^2y}{dx^2}[/tex] at x = 2?"

My initial calculation lead to 12, but then I guessed 219 as the answer and it was correct. Would any kind soul please explain why the answer would be 219? Thank you so much!

Exponential Property [Rewrite]: [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]
Exponential Property [Root Rewrite]: [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

Step-by-step explanation:

We are given the following and are trying to find the second derivative at x = 2:

[tex]\displaystyle f(2) = 2[/tex]

[tex]\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}[/tex]

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

[tex]\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}[/tex]

When we differentiate this, we must follow the Chain Rule: [tex]\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big][/tex]

Use the Basic Power Rule:

[tex]\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')[/tex]

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

[tex]\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big][/tex]

Simplifying it, we have:

[tex]\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big][/tex]

We can rewrite the 2nd derivative using exponential rules:

[tex]\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}[/tex]

To evaluate the 2nd derivative at x = 2, simply substitute in x = 2 and the value f(2) = 2 into it:

[tex]\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}[/tex]

When we evaluate this using order of operations, we should obtain our answer:

[tex]\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219[/tex]

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation