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Calculus

How to Do Implicit Differentiation? A Step-by-Step Guide With Examples

01.05.2022 • 5 min read

Rachel McLean

Subject Matter Expert

This article is a brief guide on how to do implicit differentiation. Learn how to find the implicit derivative, determine the difference between implicit and explicit differentiation, and practice finding implicit derivatives with some examples.

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In This Article

  1. What Is Implicit Differentiation?

  2. Implicit vs. Explicit Differentiation

  3. How to Do Implicit Differentiation

  4. Implicit Differentiation Example Problems

What Is Implicit Differentiation?

How can we find the derivative when xx and yy are on the same side of the equation and we can’t solve for xx or yy? 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 xx by treating yy as a function of xx.

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

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=3x2+5y = 3x^2 + 5
y=1x10y = \frac{1}{x}-10
x=2sin(θ)+2cos(θ)x = 2\sin{(\theta)}+ 2\cos{(\theta)}

Implicit functions look like this, for example:

x2+y2=16x^2+y^2=16
3x3+6xy+2y=423x^3+6xy +2y = 42
sin(xy)+cos(xy)=x\sin{(xy)} + \cos{(xy)} = x

What’s the key difference between these two types of functions? We solve an explicit function for yy in terms of xx. This means that we write an explicit function in terms of the independent variable. In an implicit function, the xx and yy 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 yy as a function of xx.

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 xx.

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

  2. Solve for dydx\frac{dy}{dx}.

Treating yy as an implicit function of xx 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:

ddxf(g(x))=f(g(x))g(x)\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 yy, the derivative of the “inside” function yy is simply dydx\frac{dy}{dx}.

So, while taking the derivative of each term that involves yy, we need to multiply by dydx\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 yy, 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

ddxxn=nxn1\frac{d}{dx}x^n = nx^{n-1}

Chain Rule

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

Product Rule

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

Quotient Rule

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

Sum/Difference Rule

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

Constant Rule

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

Trigonometry Rules

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

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

ddx(tan(x))=sec2(x)\frac{d}{dx}(\tan{(x)}) = \sec ^2 (x)

Logarithmic and Exponential Rules

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

ddx(ex)=ex\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 x2+y2=16x^2 + y^2 = 16.

Solution:

The first step is to differentiate both sides with respect to xx. 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.

ddx(x2+y2)=ddx(16)\frac{d}{dx}(x^2+y^2) = \frac{d}{dx}(16)
ddx(x2)+ddx(y2)=0\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 yy is a function of xx – we'll need to use the Chain Rule too.

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

2x+2ydydx=02x + 2y\frac{dy}{dx} = 0
2ydydx=2x2y\frac{dy}{dx} = -2x
dydx=2x2y\frac{dy}{dx} = \frac{-2x}{2y}
dydx=xy\frac{dy}{dx} = \frac{-x}{y}

Problem 2

Differentiate 3x3+6xy+2y=423x^3 + 6xy + 2y = 42.

Solution:

The first step is to differentiate both sides with respect to xx. 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.

ddx(3x3+6xy+2y)=ddx(42)\frac{d}{dx}(3x^3 + 6xy + 2y) = \frac{d}{dx}(42)
ddx(3x3)+ddx(6xy)+ddx(2y)=ddx(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 dydx\frac{dy}{dx}. We can use factoring to help with this part by factoring out dydx\frac{dy}{dx}.

9x2+6xdydx+6y+2dydx=09x^2 + 6x\frac{dy}{dx}+ 6y + 2\frac{dy}{dx} = 0
6xdydx+2dydx=9x26y6x\frac{dy}{dx} + 2\frac{dy}{dx} = -9x^2 - 6y
dydx(6x+2)=9x26y\frac{dy}{dx}(6x+2) = -9x^2 - 6y
dydx=9x26y6x+2\frac{dy}{dx} = \frac{-9x^2 – 6y}{6x + 2}

Problem 3

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

Solution:

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

ddx(sin(xy)+cos(xy))=ddxx\frac{d}{dx}(\sin{(xy)} + \cos{(xy)}) = \frac{d}{dx}x
ddx(sin(xy))+ddx(cos(xy))=1\frac{d}{dx}(\sin{(xy)}) + \frac{d}{dx}(\cos{(xy)}) = 1
cos(xy)(xdydx+y)sin(xy)(xdydx+y)=1\cos{(xy)}(x\frac{dy}{dx} + y) - \sin{(xy)}(x\frac{dy}{dx} + y) = 1

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

(xdydx+y)(cos(xy)sin(xy))=1(x\frac{dy}{dx} + y)(\cos{(xy)} - \sin{(xy)}) = 1
xdydx+y=1cos(xy)sin(xy)x\frac{dy}{dx} + y = \frac{1}{\cos{(xy)} - \sin{(xy)}}
xdydx=1cos(xy)sin(xy)yx\frac{dy}{dx} = \frac{1}{\cos{(xy)} - \sin{(xy)}} – y
dydx=1cos(xy)sin(xy)yx\frac{dy}{dx} = \frac{\frac{1}{\cos{(xy)} - \sin{(xy)}} – y }{x}
dydx=1xcos(xy)xsin(xy)yx\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 yy. In that case, you would treat xx as a function of yy, and use the Chain Rule on terms that include xx. Then, you solve for dxdy\frac{dx}{dy}.

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