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Implicit Differentiation
  • Last Updated : 25 Jan, 2021

Implicit differentiation is a method that makes use of the chain rule to differentiate implicitly defined functions. It is generally not easy to find the function explicitly and then differentiate. Instead, we can totally differentiate f(x, y) and solve the rest of the equation to find the value of dy/dx.  Even when it is possible to explicitly solve the original equation, the formula resulting from total differentiation is, in general, much simpler and easier to use. Before jumping to the topic let’s have a look at a sample problem. This will help you to understand the concepts very easily. 

Question: Find the equation for the slope of the tangent at any point on the circle centered at the origin?

Solution:

To solve this problem we need to find the equation of the function in terms of x (i.e. y = f(x)) and then find f'(x). Let us first write the general equation for a circle.

x2 + y2 = r2



In this form of the equation, the function is expressed in terms of both y and x. This function is known as implicit function as y is not explicitly defined as a function of x i.e. this function is not in the form y = f(x). Let’s find the derivative explicitly. To solve this explicitly,

  • Solve the equation for y
  • Differentiate the function
  • Substitute back y

In this case, 

x2 + y2 = r2

Subtract x2 from both sides:

y2 = r2 – x2

Taking square root on both sides:

y = ± \sqrt{r^2 − x^2}

Case 1:



y =  +\sqrt{r^2 − x^2}

=> y = (r2 x2)1/2

Derivative by Chain Rule: 

\frac{dy}{dx} =\frac{1}{2}(r^2 − x^2)^{−½}(−2x)\\ \frac{dy}{dx} =-x(r^2 − x^2)^{−½}

Since, y = +\sqrt{r^2 − x^2} \\

=> \ \frac{dy}{dx} =\frac{-x}{y}

Case 2:

y =  -\sqrt{r^2 − x^2} \\=> \ y=-({r^2 − x^2})^{1/2} \\

Derivative by Chain Rule: 

\frac{dy}{dx} =\frac{-1}{2}(r^2 − x^2)^{−½}(−2x)\\ \frac{dy}{dx} =x(r^2 − x^2)^{−½}

Since, y =  -\sqrt{r^2 − x^2} \\

\frac{dy}{dx} =\frac{x}{-y}

In either case, the answer is same i.e  dy/dx = – x/y

But our main intension in this article would be to try to solve the problem without explicitly finding the value of f(x). In this case finding the function f(x) was quite easy but it wouldn’t be the same case elsewhere.

Basic Prerequisite

  • Chain Rule: The chain rule is a formula to compute the derivative of a composite function. That is, if f and g are differentiable functions, then the chain rule expresses the derivative of their composite f ∘ g as : f(g(x))’ = f'(g(x)) * g'(x)
  • Implicit vs Explicit function: A function can be explicit or implicit:
    • Explicit: y = f(x)  e.g. y = x2
    • Implicit: f(x, y) = 0 e.g. y + x2 = 5

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

Method to Solve Implicit Differentiation

  • Differentiate both sides of the equation with respect to x.
  • Follow the rules of differentiation.
  • Use the chain rule to differentiate expressions involving y.
  • Solve the equation for dy/dx.

Now, we will solve the above question using implicit differentiation.

Solution:

Given above equation is: 

x2 + y2 = r2

Step 1: Differentiate both sides wrt to x and follow the differentiatopn

\frac{d}{dx}({x^2}+y^2) =\frac{d}{dx}r^2

Step 2: Using the chain rule

2x+2y\frac{dy}{dx}=0

Step 3: Simplify the equation

2y\frac{dy}{dx}=-2x \\ => \frac{dy}{dx}=\frac{-x}{y}

We got the same expected answer as in case of explicit differentiation.

We have successfully shown that implicit and explicit differentiation gives the same result. Let’s look at some more examples to understand the concept clearly. Sometimes the implicit way works where the explicit way is hard or impossible. In all the below examples we have to find the value of dy/dx.

Examples

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

Solution: 

Using explicit differentiation:

y + x + 5 = 0

=> y = -(x + 5)

\begin{aligned}\\ => \frac{dy}{dx}&=-(1+0) \\=> \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: 

Given equation:

y5 – y = x

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

Example 3: Find the derivative of 10x4 – 18xy2 + 10y3 = 48?

Solution: 

Given equation:

10x4 – 18xy2 + 10y3 = 48

Differentiating both sides wrt x

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

(the differentiation of term xy2 is explained below)

On simplifying we get:

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: (fg)’ = 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 equation:

x4 + 2y2 = 8

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

Example 5: Find the derivative of y = sin-1(x)?

Solution: 

Given equation:

y = sin-1(x)

=> sin y = x

\begin{aligned} \frac{d}{dx}sin (y)&= \frac{d}{dx}x \\ cos (y)\frac{dy}{dx} &= 1 \\ \frac{dy}{dx}&=\frac{1}{cos(y)} \\ \end{aligned}

We can simplify it more by using the below observation:

sin2(y) + cos2(y)

We know,

sin(y) = x

=> x2 + cos2(y) = 1

=> cos2(y) = 1 – x2

=> cos(y) = ​​\sqrt{1-x^2}

Substituting the value, we get

\begin{aligned} \frac{dy}{dx}&=\frac{1}{cos(y)} \\ \frac{dy}{dx}&=\frac{1}{\sqrt{1-x^2}} \end{aligned}

Summary

To find the derivative implicitly (useful when a function can’t easily be solved for y):

  • Differentiate with respect to x
  • Collect all the dy/dx on one side
  • Solve for dy/dx

Implicit differentiation can help us solve inverse functions. To derive an inverse function, restate it without the inverse then use Implicit differentiation.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

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