AOS4 Topic 3: Implicit Differentiation

Definition:

Implicit differentiation is a technique used in calculus to find the derivative of an implicitly defined function. An implicitly defined function is one where the dependent variable is not expressed explicitly in terms of the independent variable. Instead, it is defined implicitly through an equation involving both variables.

How It Works

  1. Implicit Functions: An implicit function is typically expressed as an equation involving both the dependent and independent variables. For example: \(x^2 + y^2 = 25\).
  2. Derivatives: To find the derivative of \(y\) with respect to \(x\), you differentiate both sides of the equation implicitly with respect to \(x\).
  3. Chain Rule: Since \(y\) is a function of \(x\), the chain rule is applied when differentiating terms involving \(y\).

Steps

  1. Differentiate Both Sides: Differentiate each term of the equation with respect to \(x\).
  2. Isolate \(y'\): Solve for \(y'\) (the derivative of \(y\) with respect to \(x\).

Example

Given \( x^2 + y^2 = 25 \), differentiate implicitly with respect to \( x \):

\[ 2x + 2yy' = 0 \]

From this point, you can solve for \( y' \) to get the derivative of \( y \) with respect to \( x \).

Types of Implicit Differentiation

  1. Basic Implicit Differentiation: Involves differentiating simple implicit functions.
  2. Higher-order Derivatives: It's possible to find higher-order derivatives using implicit differentiation by repeating the process.
  3. Implicit Functions with Multiple Variables: Sometimes functions are implicitly defined with more than one variable. In such cases, partial derivatives are used.
  4. Inverse Functions: Implicit differentiation can also be used to find the derivative of inverse functions implicitly defined.
  5. Parametric Equations: Implicit differentiation can be applied to parametric equations to find derivatives.
  6. Inverse Trigonometric Functions: Implicit differentiation is often applied to equations involving inverse trigonometric functions.

This involves differentiating simple implicit functions.

Example: Consider the equation \(x^2 + y^2 = 25\). We want to find \(\frac{dy}{dx}\).

\[ \begin{align*} \frac{d}{dx}(x^2 + y^2) &= \frac{d}{dx}(25) \\ 2x + 2y \frac{dy}{dx} &= 0 \\ 2y \frac{dy}{dx} &= -2x \\ \frac{dy}{dx} &= -\frac{x}{y} \end{align*} \]

Higher-order Derivatives

It's possible to find higher-order derivatives using implicit differentiation by repeating the process.

Example: Given \(x^2 + y^2 = 25\), find \(\frac{d^2y}{dx^2}\).

\[ \begin{align*} \frac{d}{dx}(2y \frac{dy}{dx}) &= \frac{d}{dx}(-2x) \\ 2\frac{dy}{dx}\frac{d^2y}{dx^2} + 2y \frac{d^2y}{dx^2} &= -2 \\ \frac{d^2y}{dx^2}(2\frac{dy}{dx} + 2y) &= -2 \\ \frac{d^2y}{dx^2} &= -\frac{2}{2\frac{dy}{dx} + 2y} \end{align*} \]

Inverse Functions

Implicit differentiation can also be used to find the derivative of inverse functions implicitly defined.

Example:

Given \(x^3 + y^3 = 6xy\), find \(\frac{d^2y}{dx^2}\). \begin{align*} \frac{d}{dx}(x^3 + y^3) &= \frac{d}{dx}(6xy) \\ 3x^2 + 3y^2 \frac{dy}{dx} &= 6y + 6x \frac{dy}{dx} \\ 3x^2 - 6x \frac{dy}{dx} &= 6y - 3y^2 \frac{dy}{dx} \\ 6x \frac{dy}{dx} - 3y^2 \frac{dy}{dx} &= 3x^2 - 6y \\ \frac{dy}{dx}(6x - 3y^2) &= 3x^2 - 6y \\ \frac{dy}{dx} &= \frac{3x^2 - 6y}{6x - 3y^2} \\ \end{align*}

Parametric Equations

Implicit differentiation can be applied to parametric equations to find derivatives.

Example: Given the parametric equations \(x = t^2 + 1\) and \(y = t^3 - t\), find \(\frac{dy}{dx}\).

\[ \begin{align*} \frac{d}{dt}(t^2 + 1) &= \frac{d}{dt}(t^3 - t) \\ 2t &= 3t^2 - 1 \\ \frac{dy}{dx} &= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2 - 1}{2t} \end{align*} \]

Inverse Trigonometric Functions

Implicit differentiation is often applied to equations involving inverse trigonometric functions.

Example: Given \(\sin(xy) = x^2 + y^2\), find \(\frac{dy}{dx}\).

\[ \begin{align*} \frac{d}{dx}(\sin(xy)) &= \frac{d}{dx}(x^2 + y^2) \\ \cos(xy) \left(y + x \frac{dy}{dx}\right) &= 2x + 2y \frac{dy}{dx} \\ \cos(xy)y + \cos(xy)x \frac{dy}{dx} &= 2x + 2y \frac{dy}{dx} \\ \cos(xy)x \frac{dy}{dx} - 2y \frac{dy}{dx} &= 2x - \cos(xy)y \\ \frac{dy}{dx}(x\cos(xy) - 2y) &= 2x - \cos(xy)y \\ \frac{dy}{dx} &= \frac{2x - \cos(xy)y}{x\cos(xy) - 2y} \end{align*} \]

These examples illustrate various applications of implicit differentiation and how it can be used to find derivatives in different scenarios.


Watch this video by Professor Leonard for a deep understanding


Practice using the simulation below

  • Created with GeoGebra®, by Ravinder Kumar, Link

Example 1

Find \(\frac{dy}{dx}\) by implicit differentiation for the equation \(xy = 2x + 1\).

Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(xy) = \frac{d}{dx}(2x + 1) \] \[ \frac{d}{dx}(xy) = 2 \]
Use the product rule on the left-hand side:
\[ y + x \frac{dy}{dx} = 2 \] \[ \frac{dy}{dx} = \frac{2 - y}{x} \]

Example 2

Find \( \frac{dy}{dx} \) if \( x^2 + y^2 = 1 \).

Solution:
Note that \(x^2 + y^2 = 1\) leads to \(y = \pm \sqrt{1 - x^2}\) or \(x = \pm \sqrt{1 - y^2}\). So \(y\) is not a function of \(x\), and \(x\) is not a function of \(y\). Implicit differentiation should be used. Since \(x^2 + y^2 = 1\) is the unit circle, we can also find the derivative geometrically.
Method 1: Using geometry
Let \(P(x, y)\) be a point on the unit circle with \(x \neq 0\). The gradient of \(OP\) is \(\frac{rise}{run} = \frac{y}{x}\). Since the radius is perpendicular to the tangent for a circle, the gradient of the tangent is \(-\frac{x}{y}\), provided \(y \neq 0\). That is, \(\frac{dy}{dx} = -\frac{x}{y}\).
From the graph, when \(y = 0\), the tangents are parallel to the y-axis, hence \(\frac{dy}{dx}\) is not defined.

Method 2: Using implicit differentiation
\(x^2 + y^2 = 1\)
\(2x + 2y \frac{dy}{dx} = 0\) (differentiate both sides with respect to \(x\))
\(2y \frac{dy}{dx} \therefore = -2x\)
\(\frac{dy}{dx} = -\frac{x}{y}\)
\(\therefore\) for \(y \neq 0\)

Example 3

Given \(xy - y - x^2 = 0\), find \(\frac{dy}{dx}\).

Solution
Method 1: Expressing y as a function of x
\(xy - y - x^2 = 0\)
\(y(x - 1) = x^2\)
\(y = \frac{x^2}{x - 1}\)
\(y = x + 1 + \frac{1}{x - 1}\)
Therefore for \(x \neq 1\)
\(\frac{dy}{dx} = 1 - \frac{1}{(x - 1)^2}\)
Hence
\(= \frac{(x - 1)^2 - 1}{(x - 1)^2}\)
\(= \frac{x^2 - 2x}{(x - 1)^2}\)
for \(x \neq 1\)

Method 2: Using implicit differentiation
\(xy - y - x^2 = 0\)
\(\frac{d}{dx}(xy) - \frac{dy}{dx} - \frac{d}{dx}(x^2) = \frac{d}{dx}(0)\)
\(\therefore\) (differentiate both sides with respect to \(x\))
\(x \cdot \frac{dy}{dx} + y \cdot 1 - \frac{dy}{dx} - 2x = 0\) (product rule)
\(x \frac{dy}{dx} - \frac{dy}{dx} = 2x - y \frac{dy}{dx}\)
\(\frac{dy}{dx} = \frac{2x - y}{x - 1}\)
\(\therefore\) for \(x \neq 1\)
This can be checked, by substitution of \(y = \frac{x^2}{x - 1}\), to confirm that the results are ident

Example 4

Consider the curve with equation \(2x^2 - 2xy + y^2 = 5\).
\(a\). Find \(\frac{dy}{dx}\).
\(b\). Find the gradient of the tangent to the curve at the point (1, 3).

Solution
a. Neither \(x\) nor \(y\) can be expressed as a function, so implicit differentiation must be used.
\(2x^2 - 2xy + y^2 = 5\)
\(\frac{d}{dx}(2x^2) - \frac{d}{dx}(2xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(5)\)
\(4x - \left(2x \cdot \frac{dy}{dx} + y \cdot 2\right) + 2y \frac{dy}{dx} = 0\) (by the product and chain rules)
\(4x - 2x \frac{dy}{dx} - 2y + 2y \frac{dy}{dx} = 0\)
\(2y \frac{dy}{dx} - 2x \frac{dy}{dx} = 2y - 4x \frac{dy}{dx}\)
\(\frac{2y - 2x}{2y - 2x} = \frac{2y - 4x}{2y - 2x}\)
\(\therefore = \frac{y - 2x}{y - x}\) for \(x \neq y\)

b. When \(x = 1\) and \(y = 3\), the gradient is \(\frac{3 - 2}{3 - 1} = 1\)

Example 5

Differentiate the implicit function \( y^2 + x^3 - y^3 + 6 = 3y \) with respect to \( x \).

Solution
Suppose we want to differentiate the implicit function \( y^2 + x^3 - y^3 + 6 = 3y \) with respect to \( x \).
We differentiate each term with respect to \( x \):
\[ \frac{d}{dx}(y^2) + \frac{d}{dx}(x^3) - \frac{d}{dx}(y^3) + \frac{d}{dx}(6) = \frac{d}{dx}(3y) \]
Differentiating functions of \( x \) with respect to \( x \) is straightforward. But when differentiating a function of \( y \) with respect to \( x \), we must remember the chain rule. We find:
\[ \frac{d}{dy}(y^2) \times \frac{dy}{dx} + 3x^2 - \frac{d}{dy}(y^3) \times \frac{dy}{dx} + 0 = \frac{d}{dy}(3y) \times \frac{dy}{dx} \]
That is:
\[ 2y \frac{dy}{dx} + 3x^2 - 3y^2 \frac{dy}{dx} = 3 \frac{dy}{dx} \]

We rearrange this to collect all terms involving \( \frac{dy}{dx} \) together:
\[ 3x^2 = 3 \frac{dy}{dx} - 2y \frac{dy}{dx} + 3y^2 \frac{dy}{dx} \]
Then:
\[ 3x^2 = (3 - 2y + 3y^2) \frac{dy}{dx} \]
So, finally:
\[ \frac{dy}{dx} = \frac{3x^2}{3 - 2y + 3y^2} \]
This is our expression for \( \frac{dy}{dx} \).

Example 6

Differentiate, with respect to \( x \), the implicit function \( \sin y + x^2y^3 - \cos x = 2y \).

Solution:
As before, we differentiate each term with respect to \( x \):
\[ \frac{d}{dx} (\sin y) + \frac{d}{dx} (x^2y^3) - \frac{d}{dx} (\cos x) = \frac{d}{dx} (2y) \]
Recognize that the second term is a product and we will need the product rule. We will also use the chain rule to differentiate the functions of \( y \). We find:
\[ \frac{d}{dy} (\sin y) \times \frac{dy}{dx} + x^2 \frac{d}{dx} (y^3) + y^3 \frac{d}{dx} (x^2) + \sin x = \frac{d}{dy} (2y) \times \frac{dy}{dx} \]
So that:
\[ \cos y \frac{dy}{dx} + x^2 \frac{d}{dy} (y^3) \frac{dy}{dx} + y^3 \cdot 2x + \sin x = \]

Tidying this up gives:
\(\cos y \frac{dy}{dx} + x^2 \cdot 3y^2 \frac{dy}{dx} + 2xy^3 + \sin x = 2 \frac{dy}{dx}\)
We now start to collect together terms involving \( \frac{dy}{dx} \):
\(2xy^3 + \sin x = 2 \frac{dy}{dx} - \cos y \frac{dy}{dx} - 3x^2 y^2 \frac{dy}{dx}\)
\(2xy^3 + \sin x = (2 - \cos y - 3x^2 y^2) \frac{dy}{dx}\)
So that, finally:
\(\frac{dy}{dx} = \frac{2xy^3 + \sin x}{2 - \cos y - 3x^2y^2}\)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

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