8. Differentiation of Implicit Functions

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`

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!

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:

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:

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)`