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 dydx.
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 dydx if y⁴ + x⁵ − 7x² − 5x^-1 = 0.

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:

ddxy=dydx
ddxy2=2ydydx
ddxy3=3y2dydx

It follows that:

ddxy4=4y3dydx

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

x⁵ − 7x² − 5x^-1

This is just ordinary differentiation:

ddx(x5−7x2−5x−1)=5x4−14x+5x−2

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

ddx(0)=0

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

4y3dydx+5x4−14x+5x−2=0

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

dydx=−5x4+14x−5x−24y3

Example 2

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

2y + 5 − x² − y³ = 0.

Working left to right, we have:
Derivative of 2y:

ddx2y=2dydx

Derivative of 5 is 0.
Derivative of x² is 2x.
Derivative of y³:

ddxy3=3y2dydx

Putting it together, implicit differentiation gives us:

2dydx−2x−3y2dydx=0

Collecting like terms gives:

(2−3y2)dydx=2x

So

dydx=2x2−3y2

Now, when x=2 and y=−1,

dydx=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−x2−y3=0

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