Question

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?

Answer 1

Answer:

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

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. 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.

1. Chain Rule [Basic Power Rule]:                                                                     $\displaystyle \frac{d^2y}{dx^2} = \frac{1}{2}(4y^2 + 7x^2)^{\frac{1}{2} - 1} \cdot \frac{dy}{dx} (4y^2 + 7x^2)$
2. [2nd Derivative] Simplify:                                                                               $\displaystyle \frac{d^2y}{dx^2} = \frac{1}{2}(4y^2 + 7x^2)^{-\frac{1}{2}} \cdot \frac{dy}{dx} (4y^2 + 7x^2)$
3. [2nd Derivative - Chain Rule] Basic Power Rule:                                         $\displaystyle \frac{d^2y}{dx^2} = \frac{1}{2}(4y^2 + 7x^2)^{-\frac{1}{2}} \cdot (0 + 2 \cdot 7x^{2 - 1})$
4. [2nd Derivative] Simplify:                                                                               $\displaystyle \frac{d^2y}{dx^2} = \frac{1}{2}(4y^2 + 7x^2)^{-\frac{1}{2}} \cdot 2 \cdot 7x$
5. [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.

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