In This Article

What Is Implicit Differentiation?

Implicit vs. Explicit Differentiation

How to Do Implicit Differentiation

Implicit Differentiation Example Problems

## What Is Implicit Differentiation?

How can we find the derivative when $x$ and $y$ are on the same side of the equation and we can’t solve for $x$ or $y$? Implicit differentiation is a method for finding the derivative when one or both sides of an equation have two variables that are not easily separated.

When we find the implicit derivative, we differentiate both sides of the equation with respect to the independent variable $x$ by treating $y$ as a function of $x$.

Implicit differentiation allows us to find $\frac{dy}{dx}$ without needing to solve for $y$.

##

## Implicit vs. Explicit Differentiation

What is the difference between implicit and explicit differentiation? You are probably very familiar with taking explicit derivatives. Explicit functions look like this, for example:

$y = 3x^2 + 5$

$y = \frac{1}{x}-10$

$x = 2\sin{(\theta)}+ 2\cos{(\theta)}$

Implicit functions look like this, for example:

$x^2+y^2=16$

$3x^3+6xy +2y = 42$

$\sin{(xy)} + \cos{(xy)} = x$

What’s the key difference between these two types of functions? We solve an explicit function for $y$ in terms of $x$. This means that we write an explicit function in terms of the independent variable. In an implicit function, the $x$ and $y$ variables are all mixed up, and they’re usually difficult to separate. We write an implicit function in terms of both an independent variable and a dependent variable.

What does this mean in terms of differentiation? We differentiate explicit functions normally, using the derivative rules that are familiar to you. Since implicit functions involve two mixed-up variables, we differentiate implicit functions by treating $y$ as a function of $x$.

This concept may sound familiar. Do you remember what derivative rule is necessary for taking the derivative of a composition of functions? The Chain Rule!

##

## How to Do Implicit Differentiation

Here are the two basic implicit differentiation steps. Suppose you are differentiating with respect to $x$.

Differentiate each side of the equation by treating $y$ as an implicit function of $x$. This means you need to use the Chain Rule on terms that include $y$ by multiplying by $\frac{dy}{dx}$.

Solve for $\frac{dy}{dx}$.

Treating $y$ as an implicit function of $x$ might sound complicated — but don’t worry, you’ve done this many times before whenever you use the Chain Rule! The Chain Rule is a formula for computing the derivative of a composition of functions:

$\frac{d}{dx}f(g(x)) = f’(g(x))g’(x)$

Basically, the Chain Rule says that we first take the derivative of the "outside" function, and then multiply it by the derivative of the "inside function." Since our “inside” function is $y$, the derivative of the “inside” function $y$ is simply $\frac{dy}{dx}$.

So, while taking the derivative of each term that involves $y$, we need to multiply by $\frac{dy}{dx}$.

The Chain Rule is the key to the implicit differentiation formula for success.

Besides using the Chain Rule with terms that include $y$, we can differentiate normally using the derivative rules that are already familiar to us.

For review, here are a few of the most common rules:

###
Power Rule

$\frac{d}{dx}x^n = nx^{n-1}$

###
Chain Rule

$\frac{d}{dx}f(g(x)) = f’(g(x))g’(x)$

###
Product Rule

$\frac{d}{dx}f(x) \cdot g(x) = f’(x) \cdot g(x) + f(x)\cdot g’(x)$

###
Quotient Rule

$\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{g(x)f’(x)-f(x)g’(x)}{(g(x))^2}$

###
Sum/Difference Rule

$\frac{d}{dx}(f(x) \pm g(x)) = f’(x) \pm g’(x)$

###
Constant Rule

$\frac{d}{dx}c = 0$

###
Trigonometry Rules

$\frac{d}{dx}(\sin{(x)}) = \cos{(x)}$

$\frac{d}{dx}(\cos{(x)}) = -\sin{(x)}$

$\frac{d}{dx}(\tan{(x)}) = \sec ^2 (x)$

###
Logarithmic and Exponential Rules

$\frac{d}{dx} (\ln{x}) = \frac{1}{x}$

$\frac{d}{dx}(e^x) = e^x$

## Implicit Differentiation Example Problems

To understand how to do implicit differentiation, we’ll look at some implicit differentiation examples.

###
Problem 1

Differentiate $x^2 + y^2 = 16$.

###
Solution:

The first step is to differentiate both sides with respect to $x$. Since we have a sum of functions on the left-hand side, we can use the Sum Rule. Since the right hand of the equation is a constant, we can use the Constant Rule.

$\frac{d}{dx}(x^2+y^2) = \frac{d}{dx}(16)$

$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 0$

Now, we can use the Power Rule in the first term on the right-hand side. In the second term, we can use the Power Rule, but then – because $y$ is a function of $x$ – we'll need to use the Chain Rule too.

Remember that this means multiplying by $\frac{dy}{dx}$. Then, all that’s left to do is solve for $\frac{dy}{dx}$.

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

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

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

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

### Problem 2

Differentiate $3x^3 + 6xy + 2y = 42$.

###
Solution:

The first step is to differentiate both sides with respect to $x$. Again, since we have a sum of functions on the left-hand side, we can use the Sum Rule. Since the right-hand side of the equation is a constant, we can use the Constant Rule.

$\frac{d}{dx}(3x^3 + 6xy + 2y) = \frac{d}{dx}(42)$

$\frac{d}{dx}(3x^3)+ \frac{d}{dx}(6xy)+ \frac{d}{dx}(2y) = \frac{d}{dx}(42)$

We’ll need to use the Power Rule in the first term on the right-hand side. In the second term, we need to use the Product Rule. Then, all that’s left to do is solve for $\frac{dy}{dx}$. We can use factoring to help with this part by factoring out $\frac{dy}{dx}$.

$9x^2 + 6x\frac{dy}{dx}+ 6y + 2\frac{dy}{dx} = 0$

$6x\frac{dy}{dx} + 2\frac{dy}{dx} = -9x^2 - 6y$

$\frac{dy}{dx}(6x+2) = -9x^2 - 6y$

$\frac{dy}{dx} = \frac{-9x^2 – 6y}{6x + 2}$

### Problem 3

Differentiate $\sin{(xy)} + \cos{(xy)} = x$.

###
Solution:

First we differentiate both sides with respect to $x$. We’ll use the Sum Rule. In doing so, we need to use the Chain Rule as well since $y$ is present inside the sine and cosine functions.

$\frac{d}{dx}(\sin{(xy)} + \cos{(xy)}) = \frac{d}{dx}x$

$\frac{d}{dx}(\sin{(xy)}) + \frac{d}{dx}(\cos{(xy)}) = 1$

$\cos{(xy)}(x\frac{dy}{dx} + y) - \sin{(xy)}(x\frac{dy}{dx} + y) = 1$

Now, the last step is to solve for $\frac{dy}{dx}$. We’ll do this by factoring out $(x\frac{dy}{dx} + y)$.

$(x\frac{dy}{dx} + y)(\cos{(xy)} - \sin{(xy)}) = 1$

$x\frac{dy}{dx} + y = \frac{1}{\cos{(xy)} - \sin{(xy)}}$

$x\frac{dy}{dx} = \frac{1}{\cos{(xy)} - \sin{(xy)}} – y$

$\frac{dy}{dx} = \frac{\frac{1}{\cos{(xy)} - \sin{(xy)}} – y }{x}$

$\frac{dy}{dx} = \frac{1}{x\cos{(xy)} - x\sin{(xy)}} - \frac{y}{x}$

You might also come across problems that ask you to implicitly differentiate with respect to $y$. In that case, you would treat $x$ as a function of $y$, and use the Chain Rule on terms that include $x$. Then, you solve for $\frac{dx}{dy}$.

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