Math Problem Statement

explain implicit differentiation

Solution

Implicit differentiation is a technique used to find the derivative of an equation when it's not explicitly solved for one variable in terms of another. In other words, implicit differentiation is applied when both variables (typically xx and yy) are mixed together in an equation rather than having yy isolated as a function of xx.

Steps to Implicitly Differentiate:

  1. Differentiate both sides of the equation with respect to xx: You treat yy as a function of xx (i.e., y=f(x)y = f(x)), so when differentiating terms involving yy, you apply the chain rule.

    • For example, the derivative of y2y^2 with respect to xx is: ddx(y2)=2ydydx\frac{d}{dx}(y^2) = 2y \frac{dy}{dx} This happens because you're treating yy as a function of xx, so the chain rule applies.
  2. Solve for dydx\frac{dy}{dx}: After differentiating both sides of the equation, you'll have an expression involving dydx\frac{dy}{dx}. The goal is to isolate dydx\frac{dy}{dx}, which represents the derivative of yy with respect to xx.

Example:

Consider the equation of a circle: x2+y2=25x^2 + y^2 = 25

  1. Differentiate both sides with respect to xx: ddx(x2)+ddx(y2)=ddx(25)\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)

    • ddx(x2)=2x\frac{d}{dx}(x^2) = 2x
    • ddx(y2)=2ydydx\frac{d}{dx}(y^2) = 2y \frac{dy}{dx} (using the chain rule)
    • ddx(25)=0\frac{d}{dx}(25) = 0 (since 25 is a constant)

    So the equation becomes: 2x+2ydydx=02x + 2y \frac{dy}{dx} = 0

  2. Solve for dydx\frac{dy}{dx}: 2ydydx=2x2y \frac{dy}{dx} = -2x dydx=2x2y=xy\frac{dy}{dx} = \frac{-2x}{2y} = \frac{-x}{y}

Thus, the derivative of yy with respect to xx for the equation x2+y2=25x^2 + y^2 = 25 is dydx=xy\frac{dy}{dx} = \frac{-x}{y}.

Key Points:

  • Use implicit differentiation when yy cannot be easily isolated in the equation.
  • Apply the chain rule when differentiating terms with yy.
  • After differentiation, solve for dydx\frac{dy}{dx}.

Would you like more details or examples on any part of implicit differentiation?

Here are 5 related questions to deepen your understanding:

  1. How is implicit differentiation applied to trigonometric functions?
  2. How would you differentiate x2+xy+y2=10x^2 + xy + y^2 = 10?
  3. What is the difference between explicit and implicit differentiation?
  4. Can implicit differentiation be used to find second derivatives?
  5. How does implicit differentiation apply to parametric equations?

Tip: Always remember to apply the chain rule when differentiating terms involving yy, since yy is treated as a function of xx.

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Math Problem Analysis

Mathematical Concepts

Calculus
Implicit Differentiation
Chain Rule

Formulas

\( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \)
\( \frac{dy}{dx} = \frac{-x}{y} \)

Theorems

Chain Rule

Suitable Grade Level

Grades 11-12