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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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
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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