#Derivatives of Implicit Functions

#Table of Contents

1. Introduction to Implicit Functions
2. Implicit Differentiation
3. Examples and Applications
4. Differentiating Inverse Functions
5. Practice Questions
6. Glossary
7. Summary and Key Takeaways

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#Introduction to Implicit Functions

#What is an Implicit Function?

• An equation in the form $y=f(x)$ or $x=f(y)$ is written explicitly.

• Equations involving both $x$ and $y$ are referred to as implicit functions.

• For such equations, we cannot express $y$ solely in terms of $x$ or vice versa.
• However, these equations define a relationship between $x$ and $y$.

#Implicit Differentiation

#What is Implicit Differentiation?

• Implicit differentiation is the method used to differentiate implicit functions.
• To differentiate an implicit function with respect to $x$, each term is differentiated with respect to $x$.
• For terms involving only $x$, this is straightforward.
• For terms involving $y$, we apply the chain rule.

• Once each term has been differentiated, rearrange the equation to solve for $\frac{dy}{dx}$.
• Factorize out $\frac{dy}{dx}$ if necessary.
• Substitute specific $(x, y)$ values to find the derivative at a point if required.

#How to Use Implicit Differentiation?

#Example

Differentiate $x^2 + y^2 = 4x$ implicitly.

1. Differentiate each term with respect to $x$: $$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4x)$$ $$2x + 2y \cdot \frac{dy}{dx} = 4$$

2. Rearrange to solve for $\frac{dy}{dx}$: $$2y \cdot \frac{dy}{dx} = 4 - 2x$$ $$\frac{dy}{dx} = \frac{4 - 2x}{2y}$$ $$\frac{dy}{dx} = \frac{2 - x}{y}$$

#Worked Example

A curve is defined by the equation $4x^3 + 2x + 2y^3 + 6y = 44$.

#(a) Confirm that the point $(2, 1)$ lies on the curve.

Solution: Substitute $x = 2$ and $y = 1$ into the equation: $$4(2)^3 + 2(2) + 2(1)^3 + 6(1) = 44$$ $$32 + 4 + 2 + 6 = 44$$ $$44 = 44$$ The equation is satisfied, so the point lies on the curve.

#(b) Find the value of the derivative $\frac{dy}{dx}$ at the point $(2, 1)$.

Solution: Differentiate each term with respect to $x$: $$\frac{d}{dx}(4x^3) + \frac{d}{dx}(2x) + \frac{d}{dx}(2y^3) + \frac{d}{dx}(6y) = 0$$ $$12x^2 + 2 + 6y^2 \cdot \frac{dy}{dx} + 6 \cdot \frac{dy}{dx} = 0$$

Factorize and solve for $\frac{dy}{dx}$: $$12x^2 + 2 + \frac{dy}{dx}(6y^2 + 6) = 0$$ $$\frac{dy}{dx} = \frac{-12x^2 - 2}{6y^2 + 6}$$ $$\frac{dy}{dx} = \frac{-6x^2 - 1}{3y^2 + 3}$$

Substitute $x = 2$ and $y = 1$: $$\frac{dy}{dx} = \frac{-6(2)^2 - 1}{3(1)^2 + 3}$$ $$\frac{dy}{dx} = \frac{-24 - 1}{6}$$ $$\frac{dy}{dx} = -\frac{25}{6}$$

The value of the derivative at $(2, 1)$ is $-\frac{25}{6}$.

#Examples and Applications

#Harder Implicit Differentiation Questions

• Implicit differentiation may involve additional rules:
  • Chain rule
  • Product rule
  • Quotient rule
  • Derivatives of exponentials, logarithms, and trigonometric functions

#Useful Result for Product Rule:

$$\frac{d}{dx}(xy) = y + x \frac{dy}{dx}$$

#Worked Example

Given $\frac{dy}{dx}$ for $4x^3 + 3y^2 \cdot \frac{dy}{dx} = \frac{d}{dx}(12xy)$:

Solution: Differentiate each term with respect to $x$: $$4x^3 + 3y^2 \cdot \frac{dy}{dx} = \frac{d}{dx}(12xy)$$

Use the product rule for $12xy$: $$\frac{d}{dx}(12xy) = 12y + 12x \cdot \frac{dy}{dx}$$

Combine and solve for $\frac{dy}{dx}$: $$4x^3 + 3y^2 \cdot \frac{dy}{dx} = 12y + 12x \cdot \frac{dy}{dx}$$ $$3y^2 \cdot \frac{dy}{dx} - 12x \cdot \frac{dy}{dx} = 12y - 4x^3$$ $$\frac{dy}{dx}(3y^2 - 12x) = 12y - 4x^3$$ $$\frac{dy}{dx} = \frac{12y - 4x^3}{3y^2 - 12x}$$

Equivalent answers may also be correct, e.g.: $$\frac{dy}{dx} = \frac{4x^3 - 12y}{12x - 3y^2}$$

#Differentiating Inverse Functions

#How to Differentiate Inverse Functions Using Implicit Differentiation?

• Implicit differentiation can be used for finding the derivative of inverse functions.

• Consider differentiating $y = \sin^{-1}(x)$.

  • Rewrite as $x = \sin(y)$.

• Differentiate using implicit differentiation: $$\frac{d}{dx}(x) = \frac{d}{dx}(\sin(y))$$ $$1 = \cos(y) \cdot \frac{dy}{dx}$$

• Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{1}{\cos(y)}$$

• Recall $y = \sin^{-1}(x)$: $$\frac{dy}{dx} = \frac{1}{\cos(\sin^{-1}(x))}$$

• Use the identity $\sin^2(x) + \cos^2(x) = 1$ rearranged as $\cos(x) = \sqrt{1 - \sin^2(x)}$: $$\frac{dy}{dx} = \frac{1}{\sqrt{1 - \sin^2(\sin^{-1}(x))}}$$ $$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$$

#Worked Example

Given $y = \arctan(x)$, use implicit differentiation to show that $\frac{dy}{dx} = \frac{1}{x^2 + 1}$.

Solution:

• Rewrite as $\tan(y) = x$.

• Differentiate using implicit differentiation: $$\frac{d}{dx}(\tan(y)) = \frac{d}{dx}(x)$$ $$\sec^2(y) \cdot \frac{dy}{dx} = 1$$

• Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{1}{\sec^2(y)}$$

• Recall $y = \arctan(x)$: $$\frac{dy}{dx} = \frac{1}{\sec^2(\arctan(x))}$$

• Use the identity $\tan^2(x) + 1 = \sec^2(x)$: $$\frac{dy}{dx} = \frac{1}{\tan^2(\arctan(x)) + 1} = \frac{1}{x^2 + 1}$$

#Practice Questions

Practice Question

Question 1: Differentiate $e^x + e^y = 1$ implicitly.

Question 2: Given $x^2 + 2xy + y^2 = 1$, find $\frac{dy}{dx}$.

Question 3: Verify that the point $(1, 1)$ lies on the curve $x^3 + y^3 = 1$ and find $\frac{dy}{dx}$ at that point.

#Glossary

• Implicit Function: A function where $y$ cannot be explicitly written as $y=f(x)$.
• Implicit Differentiation: A technique to find the derivative of implicit functions.
• Chain Rule: A rule to differentiate composite functions.
• Product Rule: A rule to differentiate products of two functions.
• Quotient Rule: A rule to differentiate quotients of two functions.
• Inverse Function: A function that reverses another function.

#Summary and Key Takeaways

#Summary

• Implicit functions involve both $x$ and $y$ in an equation.
• Implicit differentiation uses the chain rule to differentiate these equations.
• Rearrange the differentiated equation to solve for $\frac{dy}{dx}$.
• This method can also be used to differentiate inverse functions.

#Key Takeaways

• Understand the implicit function and how to differentiate each term.
• Apply the chain rule when differentiating $y$-terms.
• Rearrange and solve for $\frac{dy}{dx}$.
• Use implicit differentiation for inverse functions as well.

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These structured notes should help students understand and apply the concepts of implicit differentiation effectively.