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:
- 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`)
- The curve has two "arms"
- 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.`
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`
Example 2
Find the slope of the tangent at the point `(2,-1)` for the curve:
2y + 5 − x2 − y3 = 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!
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
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)`
