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.
Find the expression for `(dy)/(dx)` 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:
It follows that:
Part B: Find the derivative with respect to x of:
x5 − 7x2 − 5x-1
This is just ordinary differentiation:
On the right hand side of our expression, the derivative of zero is zero. ie
Now, combining the results of parts A, B and C:
Next, solve for dy/dx and the required expression is:
Easy to understand math videos:
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`:
Derivative of `5` is `0`.
Derivative of x2 is `2x`.
Derivative of y3:
Putting it together, implicit differentiation gives us:
Collecting like terms gives:
Now, when `x = 2` and `y = -1`,
So the slope of the tangent at `(2,-1)` is `-4`.
Let's see what we have done. We graph the curve
and graph the tangent to the curve at `(2, -1)`. We see that indeed the slope is `-4`.
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.)
To make life easy, we will break this question up into parts.
Find the derivative with respect to x of: y4
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:
Now to find `(dy)/(dx)` for the whole expression:
Working left to right, using our answers from above:
This gives us, on collecting terms:
So we have the required expression: