Implicit Differentiation

Implicit Differentiation is the process of differentiation in which we differentiate the implicit function without converting it into an explicit function. For example, we need to find the slope of a circle with an origin at 0 and a radius r. Its equation is given as x² + y² = r². 

Now, to find the slope we need to find the dy/dx of the given function, so without implicit differentiation, we have to convert this function into an explicit function i.e., y = ∓√(r² - x²) . The explicit function of this is comparatively hard to differentiate. Thus, we need to learn the implicit differentiation by which this can be very easily differentiated.

Prerequisite for Implicit Differentiation

There are some prerequisite concepts that we need to know before learning Implicit Differentiation, these prerequisites are as follows:

• Chain Rule
• Implicit and Explicit Function

When a function is not defined explicitly in terms of a single independent variable, is called implicit function. For example, y + x² = 5,  x² + y² = r², etc.

When a function is defined in terms of a single independent variable explicitly such as y = f(x), then the function is called the explicit function. For example, y = x², y = 3x+7, y = sin x, etc.

Note: Here we took only 2 variables x and y to define the implicit function. But you can have any number of variables.

How to do Implicit Differentiation

The following steps need to be followed to differentiate any implicit function.

Step 1: Follow the rules of differentiation to differentiate both sides of the equation with respect to x.

Step 2: Use the chain rule to differentiate expressions involving y.

Step 3: Solve the equation for dy/dx.

Let's consider an example for better understanding.

Example: Differentiate x² + y² = r².

Solution:

Given equation: x² + y² = r²

Step 1: Differentiate both sides wrt to x and follow the rules of differentiation.
d/dx{x² + y²} = d/dx(r²)

Step 2: Using the chain rule
2x + 2y(dy/dx) = 0

Step 3: Simplify the equation
2y(dy/dx) = -2x
⇒ dy/dx = -x/y

Thus, dy/dx = -x/y

Implicit Differentiation of Inverse Trigonometric Functions

Implicit differentiation is very useful in finding derivatives of Inverse trigonometric Functions.

Let us consider y = sin^-1(x) and we need to find its derivative, 

Take sin both sides of the equation,

sin y = sin(sin^-1(x)
⇒ sin y = x

Differentiating the above equation w.r.t x, we get

d/dx(sin y) = d/dx (x)
⇒ cos y(dy/dx) = 1
⇒ dy/dx = 1/(cos y)

Now, sin(y) = x
⇒ x² = sin²(y) 
⇒ x² + cos²(y) = 1
⇒ cos²(y) = 1 - x²
⇒ cos(y) = ​​\sqrt{1-x^2}

Substituting the value, we get
dy/dx = 1/(cos y)
dy/dx = 1/√(1 - x²)

Difference between Implicit and Explicit Differentiation

Implicit Differentiation and Explicit Differentiation are the two methods of differentiation that are used in calculus. The differences between them are explained in the table below,

Read More,

• Derivatives
• Differentiation Formulas

Implicit Differentiation Examples

Example 1: Find the derivative of y + x + 5 = 0.

Solution: 

Using Explicit Differentiation

y + x + 5 = 0
⇒ y = -(x + 5)
\begin{aligned}\\ \Rightarrow \frac{dy}{dx}& =-(1+0) \\\Rightarrow \frac{dy}{dx}&=-1 \end{aligned}

Using Implicit Differentiation

y + x + 5 = 0

Differentiating both sides wrt x
\frac{dy}{dx}+1+0=0

Isolate dy/dx
\frac{dy}{dx}=-1

Example 2: Find the derivative of y⁵ - y = x.

Solution: 

y⁵ - y = x

Differentiating the above equation with respect to x, we get

\begin{aligned} & \Rightarrow 5y^4\frac{dy}{dx}-\frac{dy}{dx}&=1 \\ & \Rightarrow (5y^4-1)\frac{dy}{dx}&=1 \\ & \Rightarrow \frac{dy}{dx}=\frac{1}{5y^4-1} \end{aligned}

Example 3: Find the derivative of 10x⁴ - 18xy² + 10y³ = 48.

Solution: 

Given, 10x⁴ - 18xy² + 10y³ = 48

Differentiating both sides w.r.t x

10(4x^3) − 18(x.2y\frac{dy}{dx} + y^2) + 10(3y^2 \frac{dy}{dx}) = 0
⇒ 40x^3 − 36xy \frac{dy}{dx} − 18y^2 + 30y^2 \frac{dy}{dx} = 0

Keeping all the terms involving dy/dx on left and rest terms on right side of equation

−36xy \frac{dy}{dx} + 30y^2 \frac{dy}{dx} = −40x^3+ 18y^2
⇒ (30y^2−36xy)\frac{dy}{dx}= 18y^2 − 40x^3

Dividing both sides by 2
⇒ (15y^2−18xy) \frac{dy}{dx}=9y^2 − 20x^3

Finally Isolate dy/dx
⇒ \bold{\frac{dy}{dx}=\frac{9y^2 − 20x^3}{15y^2−18xy}}

For the term xy² we used the Product Rule: (f.g)’ = f.g’ + f’.g
\begin{aligned} (xy^2)’  &= x(y^2)’ + (x)’y^2 \\ &=x({2y\frac{dy}{dx}})+y^2 \\ &=2xy\frac{dy}{dx}+y^2 \end{aligned}

Example 4: Find the derivative of x⁴ + 2y² = 8.

Solution: 

Given, x⁴ + 2y² = 8

\begin{aligned}  4x^3+4y\frac{dy}{dx}&=0 \\ \frac{dy}{dx}&=\frac{-x^3}{y} \end{aligned}

Also Check; Implicit Differentiation Advance Examples.