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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:

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

Find the expression for `(dy)/(dx)` if y4 + x5 − 7x2 − 5x-1 = 0.

Answer

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`

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Continues below

Example 2

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

2y + 5 − x2y3 = 0.

Answer

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!

implicit graph

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Example 3 (Involves Product Rule)

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

y4 + 2x2y2 + 6x2 = 7

(This is the example given at the top of this page.)

Answer

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)`

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