Implicit Differentiation

Finding the derivative when you can’t solve for y

What are you trying to imply?

You may like to read Introduction to Derivatives and Derivative Rules first.

Implicit vs Explicit

A function can be explicit or implicit:

Explicit: "y = some function of x". When we know x we can calculate y directly.

Implicit: "some function of y and x equals something else". Knowing x does not lead directly to y.

Example: A Circle

Explicit Form   Implicit Form
y = ± √ (r2 − x2)   x2 + y2 = r2
In this form, y is expressed
as a function of x.
  In this form, the function is
expressed in terms of both y and x.

graph x^2 + y^2 = 9, a circle
The graph of x2 + y2 = 32

How to do Implicit Differentiation

Example: x2 + y2 = r2

Differentiate with respect to x:

d dx (x2) + d dx (y2) = d dx (r2)

Let's solve each term:

Use the Power Rule: d dx (x2) = 2x
Use the Chain Rule (explained below): d dx (y2) = 2y dy dx
r2 is a constant, so its derivative is 0: d dx (r2) = 0

Which gives us:

2x + 2y dy dx = 0

Collect all the dy dx on one side

y dy dx = −x

Solve for dy dx :

dy dx = −x y

The Chain Rule Using dy dx

Let's look more closely at how d dx (y2) becomes 2y dy dx

The Chain Rule says:

du dx = du dy dy dx

Substitute in u = y2:

d dx (y2) = d dy (y2) dy dx

And then:

d dx (y2) = 2y dy dx

Basically, all we did was differentiate with respect to y and multiply by dy dx

Another common notation is to use to mean d dx

The Chain Rule Using 

The Chain Rule can also be written using notation:

f(g(x))’ = f’(g(x))g’(x)

g(x) is our function "y", so:

f(y)’ = f’(y)y’

f(y) = y2, so f(y) = 2y:

f(y)’ = 2yy’

or alternatively: f(y)’ = 2y dy dx

Again, all we did was differentiate with respect to y and multiply by dy dx


Let's also find the derivative using the explicit form of the equation.

Example: x2 + y2 = r2

Subtract x2 from both sides:y2 = r2 − x2
Square root:y = ±√(r2 − x2
Let's do just the positive: y = √(r2 − x2
As a power: y = (r2 − x2)½
Derivative (Chain Rule):y =½(r2 − x2)−½(2x)
Simplify:y = −x(r2 − x2)−½
Simplify more:y = −x (r2 − x2)½
Now, because y = (r2 − x2)½y = −x/y

We get the same result this way!

You can try taking the derivative of the negative term yourself.

Chain Rule Again!

Yes, we used the Chain Rule again. Like this (note different letters, but same rule):

dy dx = dy df df dx

Substitute in f = (r2 − x2):

d dx (f½) = d df (f½) d dx (r2 − x2)


d dx (f½) = ½(f−½) (−2x)

And substitute back f = (r2 − x2):

d dx (r2 − x2)½ = ½((r2 − x2)−½) (−2x)

And we simplified from there.

Using The Derivative

OK, so why find the derivative y’ = −x/y ?

Well, for example, we can find the slope of a tangent line.

Example: what is the slope of a circle centered at the origin with a radius of 5 at the point (3, 4)?

graph x^2 + y^2 = 25 with tangent line

No problem, just substitute it into our equation:

dy dx = −x/y

dy dx = −3/4

And for bonus, the equation for the tangent line is:

y = −3/4 x + 25/4

Another Example

Sometimes the implicit way works where the explicit way is hard or impossible.

Example: 10x4 − 18xy2 + 10y3 = 48

How do we solve for y? We don't have to!

  • First, differentiate with respect to x (use the Product Rule for the xy2 term).
  • Then move all dy/dx terms to the left side.
  • Solve for dy/dx

Like this:

Start with:10x4 − 18xy2 + 10y3 = 48
Derivative:10 (4x3) − 18(x(2y dy dx ) + y2) + 10(3y2 dy dx ) = 0

(the middle term is explained
in "Product Rule" below)

Simplify:40x3 − 36xy dy dx − 18y2 + 30y2 dy dx = 0
dy dx on left:−36xy dy dx + 30y2dy dx = −40x3 + 18y2
Simplify :(30y2−36xy) dy dx = 18y2 − 40x3
Simplify :3(5y2−6xy) dy dx = 9y2 − 20x3

And we get:

dydx = 9y2 − 20x33(5y2 − 6xy)


Product Rule

For the middle term we used the Product Rule: (fg)’ = f g’ + f’ g

(xy2)’ = x(y2)’ + (x)’y2
 = x(2y dy dx ) + y2

Because (y2)’  = 2y dy dx (we worked that out in a previous example)

Oh, and dxdx = 1, in other words x’ = 1

Inverse Functions

Implicit differentiation can help us solve inverse functions.

The general pattern is:

As a final step we can try to simplify more by substituting the original equation.

An example will help:

Example: the inverse sine function y = sin−1(x)

Start with:y = sin−1(x)
In non−inverse mode:x = sin(y)
Derivative: d dx (x) = d dx sin(y)
 1 = cos(y) dy dx
Put dy dx on left: dy dx = 1 cos(y)

We can also go one step further using the Pythagorean identity:

sin2 y + cos2 y = 1

cos y = √(1 − sin2 y )

And, because sin(y) = x (from above!), we get:

cos y = √(1 − x2)

Which leads to:

dy dx = 1 √(1 − x2)

Example: the derivative of square root √x

Start with:y = √x
So:y2 = x
Derivative:2y dy dx = 1
Simplify: dy dx = 1 2y
Because y = √x: dy dx = 1 2√x

Note: this is the same answer we get using the Power Rule:

Start with:y = √x
As a power:y = x½
Power Rule d dx xn = nxn−1: dy dx = (½)x−½
Simplify: dy dx = 1 2√x



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