Mathematics, 11.01.2021 16:10 brainbean

Let y=f(x) be a twice differentiable function such that f(1)=3 and dy/dx= (4y^2+7x^2)^1/2. What is the value of d2y/dx2 at x=1?

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Answer from: AnastasiaJauregui

$\displaystyle f''(1) = \frac{7}{\sqrt{43}}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Algebra I

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate y₁ - y coordinate m - slope

Calculus

Derivatives

Derivative Notation

Taking Derivatives with respect to x

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

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

Step-by-step explanation:

Step 1: Define

Function: y = f(x), twice differentiable.

$\displaystyle f(1) = 3\\\frac{dy}{dx} = (4y^2 + 7x^2)^{\frac{1}{2}}$

Step 2: Differentiate

Remember we are taking the derivative with respect to x.

Chain Rule [Basic Power Rule]: $\displaystyle \frac{d^2y}{dx^2} = \frac{1}{2}(4y^2 + 7x^2)^{\frac{1}{2} - 1} \cdot \frac{dy}{dx} (4y^2 + 7x^2)$ [2nd Derivative] Simplify: $\displaystyle \frac{d^2y}{dx^2} = \frac{1}{2}(4y^2 + 7x^2)^{-\frac{1}{2}} \cdot \frac{dy}{dx} (4y^2 + 7x^2)$ [2nd Derivative - Chain Rule] Basic Power Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{1}{2}(4y^2 + 7x^2)^{-\frac{1}{2}} \cdot (0 + 2 \cdot 7x^{2 - 1})$ [2nd Derivative] Simplify: $\displaystyle \frac{d^2y}{dx^2} = \frac{1}{2}(4y^2 + 7x^2)^{-\frac{1}{2}} \cdot 2 \cdot 7x$ [2nd Derivative] Simplify: $\displaystyle \frac{d^2y}{dx^2} = \frac{7x}{\sqrt{7x^2 + 4y^2}}$

Step 3: Evaluate

We are given x = 1 and f(1) = 3. This will tell us the instantaneous slope.

Substitute [2nd Deriv]: $\displaystyle f''(1) = \frac{7(1)}{\sqrt{7(1)^2 + 4(3)^2}}$ [√Radical] Exponents: $\displaystyle f''(1) = \frac{7(1)}{\sqrt{7(1) + 4(9)}}$ [Fraction] Multiply: $\displaystyle f''(1) = \frac{7}{\sqrt{7 + 36}}$ [√Radical] Add: $\displaystyle f''(1) = \frac{7}{\sqrt{43}}$

This tells us that the rate of change of the slope of the tangent line is $\displaystyle \frac{7}{\sqrt{43}}$ .

We can also write an equation for the instantaneous slope:

[Equation] $\displaystyle y - 3 = \frac{7}{\sqrt{43}}(x - 1)$

Answer from: Quest

The answer is 30 minutes to an hour

Answer from: Quest

you first to choose the points and then move from then as the slope you have

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