Search IntMath
Close

# 8. Differentiation of Implicit Functions

by M. Bourne

We meet many equations where y is not expressed explicitly in terms of x only, such as:

f(x, y) = y4 + 2x2y2 + 6x2 = 7

You can see several examples of such expressions in the Polar Graphs section.

It is usually difficult, if not impossible, to solve for y so that we can then find (dy)/(dx).

We need to be able to find derivatives of such expressions to find the rate of change of y as x changes. To do this, we need to know implicit differentiation.

Let's learn how this works in some examples.

## Example 1

We begin with the implicit function y4 + x5 − 7x2 − 5x-1 = 0.

Here is the graph of that implicit function. Observe:

1. It is not an ordinary function because there's more than one y-value for each x-value (for the regions x < -1 and 0 < x < 2)
2. The curve has two "arms"
3. The curve is vertical near x = -1 and x = 2

Find the derivative of this implicit function, and express the answer in the form dy/dx.

y4 + x5 − 7x2 − 5x-1 = 0

We see how to derive this expression one part at a time. We just derive expressions as we come to them from left to right.

(In this example we could easily express the function in terms of y only, but this is intended as a relatively simple first example.)

Part A: Find the derivative with respect to x of: y4

To differentiate this expression, we regard y as a function of x and use the power rule.

Basics: Observe the following pattern of derivatives:

d/(dx)y=(dy)/(dx)

d/(dx)y^2=2y(dy)/(dx)

d/(dx)y^3=3y^2(dy)/(dx)

It follows that:

d/(dx)y^4=4y^3(dy)/(dx)

Part B: Find the derivative with respect to x of:

x5 − 7x2 − 5x-1

This is just ordinary differentiation:

d/(dx)(x^5-7x^2-5x^-1) =5x^4-14x+5x^-2

Part C:

On the right hand side of our expression, the derivative of zero is zero. ie

d/(dx)(0)=0

Now, combining the results of parts A, B and C:

4y^3(dy)/(dx)+5x^4-14x+5x^-2=0

Next, solve for dy/dx and the required expression is:

(dy)/(dx)=(-5x^4+14x-5x^-2)/(4y^3

### Example 2

Find the slope of the tangent at the point (2,-1) for the curve:

2y + 5 − x2y3 = 0.

Working left to right, we have:

Derivative of 2y:

d/(dx)2y=2(dy)/(dx)

Derivative of 5 is 0.

Derivative of x2 is 2x.

Derivative of y3:

d/(dx)y^3=3y^2(dy)/(dx)

Putting it together, implicit differentiation gives us:

2(dy)/(dx)-2x-3y^2(dy)/(dx)=0

Collecting like terms gives:

(2-3y^2)(dy)/(dx)=2x

So

(dy)/(dx)=(2x)/(2-3y^2)

Now, when x = 2 and y = -1,

(dy)/(dx)=(2(2))/(2-3(-1)^2)

=4/-1

=-4

So the slope of the tangent at (2,-1) is -4.

Let's see what we have done. We graph the curve

2y+5-x^2-y^3=0

and graph the tangent to the curve at (2, -1). We see that indeed the slope is -4.

It works!

### Example 3 (Involves Product Rule)

Find the expression for (dy)/(dx) if:

y4 + 2x2y2 + 6x2 = 7

First, let's graph the implicit function given in the question to see what we are working with. We observe it is simply an ellipse:

Graph of y4 + 2x2y2 + 6x2 = 7

To make life easy, we will break this question up into parts.

### Part A:

Find the derivative with respect to x of: y4

d/(dx)y^4=4y^3(dy)/(dx)

### Part B:

Find the derivative with respect to x of 2x2y2

Now to find the derivative of 2x2y2 with respect to x we must recognise that it is a product.

If we let u = 2x2 and v = y2 then we have:

d/(dx)(2x^2y^2)=u(dv)/(dx)+v(du)/(dx)

=(2x^2)(2y(dy)/(dx))+ (y^2)(4x)

=4x^2y(dy)/(dx)+4xy^2

### Part C:

Now

d/(dx)6x^2=12x

and

d/(dx)(7)=0

Now to find (dy)/(dx) for the whole expression:

y^4+2x^2y^2+6x^2=7

Working left to right, using our answers from above:

[4y^3(dy)/(dx)]+[4x^2y(dy)/(dx)+4xy^2]+ [12x]=0

This gives us, on collecting terms:

(4y^3+4x^2y)(dy)/(dx)=-4xy^2-12x

So we have the required expression:

(dy)/(dx)=(-4xy^2-12x)/(4y^3+4x^2y)=(-xy^2-3x)/(y^3+x^2y)