Math Problem Statement

Find d²y/dx² in terms of x and y, given y³ = x⁴.

Solution

We are asked to find the second derivative d2ydx2\frac{d^2 y}{dx^2} in terms of xx and yy, given the equation y3=x4y^3 = x^4.

Step 1: Implicit Differentiation for the First Derivative

We start by differentiating both sides of the equation y3=x4y^3 = x^4 with respect to xx, using implicit differentiation.

ddx(y3)=ddx(x4)\frac{d}{dx} \left( y^3 \right) = \frac{d}{dx} \left( x^4 \right)

Using the chain rule for y3y^3, we get:

3y2dydx=4x33y^2 \frac{dy}{dx} = 4x^3

Solving for dydx\frac{dy}{dx}:

dydx=4x33y2\frac{dy}{dx} = \frac{4x^3}{3y^2}

Step 2: Implicit Differentiation for the Second Derivative

Now, to find d2ydx2\frac{d^2y}{dx^2}, we differentiate dydx=4x33y2\frac{dy}{dx} = \frac{4x^3}{3y^2} again with respect to xx:

ddx(dydx)=ddx(4x33y2)\frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} \left( \frac{4x^3}{3y^2} \right)

This requires using the quotient rule:

ddx(uv)=vdudxudvdxv2\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

Here, u=4x3u = 4x^3 and v=3y2v = 3y^2. Now, compute the derivatives:

  • dudx=12x2\frac{du}{dx} = 12x^2
  • dvdx=6ydydx\frac{dv}{dx} = 6y \frac{dy}{dx}, and we substitute dydx=4x33y2\frac{dy}{dx} = \frac{4x^3}{3y^2}

Thus:

dvdx=6y4x33y2=24x3y\frac{dv}{dx} = 6y \cdot \frac{4x^3}{3y^2} = \frac{24x^3}{y}

Now, apply the quotient rule:

d2ydx2=3y212x24x324x3y(3y2)2\frac{d^2y}{dx^2} = \frac{3y^2 \cdot 12x^2 - 4x^3 \cdot \frac{24x^3}{y}}{(3y^2)^2}

Simplify the numerator:

=36y2x296x6/y9y4= \frac{36y^2x^2 - 96x^6 / y}{9y^4}

Final Expression:

d2ydx2=4x2(y34x4)3y5\frac{d^2y}{dx^2} = \frac{4x^2 \left( y^3 - 4x^4 \right)}{3y^5}

This is the second derivative of yy with respect to xx in terms of xx and yy.

Would you like further details on any steps, or have any questions?

Related questions:

  1. How is implicit differentiation applied in multi-variable calculus?
  2. What are the advantages of using the quotient rule in differentiation?
  3. Can you explain how the chain rule works for implicit differentiation?
  4. How would the result change if the equation was y4=x5y^4 = x^5 instead?
  5. What physical phenomena might be modeled by an equation like y3=x4y^3 = x^4?

Tip:

Always simplify the expression as much as possible after differentiating to make interpretation and further manipulation easier.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Chain Rule
Quotient Rule

Formulas

Implicit Differentiation: d/dx (f(y)) = f'(y) * dy/dx
Quotient Rule: d/dx (u/v) = (v * du/dx - u * dv/dx) / v²

Theorems

Chain Rule
Quotient Rule

Suitable Grade Level

University Level Calculus