Implicit Differentiation and Differentiating Parametric Equations

More differentiation!

I'm going to do parametric equations first, because they're easier.

Differentiating Parametric Equations

Parametric equations are where you have separate equations for x and y in terms of a third variable, like t. For example, you could have x = 2t and y = t^2 + 1. Sometimes you can combine the two functions to make one single function in terms of either x or y. In this example, you could rearrange x = 2t to become t = (x/2), and then substitute that into the y equation to find an equation for y in terms of x. You could then differentiate that way. Sometimes, though, it's rather tiresome and annoying to do all this substitution and whatnot, which is why there's a method for directly differentiating parametric equations. It's a pretty simple method too, which is nice.

First, you find dx/dt (or, if x isn't given in terms of t, just replace the t with whatever variable x is given in terms of). Then you find dy/dt. Now, by the chain rule, dy/dx = (dy/dt)(dt/dx). But wait! you might say. We have dx/dt, but not dt/dx! Well, the handy thing is, dt/dx is simply equal to 1/(dx/dt).

For the above equations, x = 2t and y = t^2 + 1, dx/dt = 2 and dy/dt = 2t. dy/dx = (dy/dt)(dt/dx) = (2t)(1/2). Therefore, dy/dx = t. If you want to find this in terms of x, just use one of your original equations to find what t is. x = 2t can be arranged to t = x/2. Therefore, dy/dx = t = x/2. You can get the same answer using the substitution method that I described above.

Differentiating Implicitly Defined Functions

Most of the time, when you see equations, they're in the form of y in terms of x or x in terms of y. But sometimes, you might get an equation where x and y are on the same side, like in x^2 + y^2 = 16 (which, by the way, is the graph of a circle, centre (0, 0) and radius 4). One way of differentiating such equations is to rearrange them and then differentiate, but sometimes this is hard or even impossible to do. That's where implicit differentiation comes in!

Implicit differentiation, however, is pretty long-winded, so if you're doing a calculator assumed part of a test and you see an implicit differentiation question only worth 1 or 2 marks, use your calculator instead! It's not worth the effort!

Basically, in implicit differentiation, you have to differentiate each term separately with respect to x (or one of the two variables in the equation). Then you have to factor out dy/dx (or dy/dt or whatever the variables are) to get your derivative.

Here's how to differentiate the different terms. Let x = whatever variable you're differentiating with respect to and y = the other variable.

Number on its own: Derivative is 0 (like always)
x, or powers of x: Differentiate as normal
y: Derivative of y is dy/dx. (dy/dx does mean "derivative of y with respect to x"...)
Powers of y: Use the chain rule as well as the above fact. y^2 becomes (2y)(dy/dx)
Mix of x and y: Separate into two groups, one with the x and one with the y. For example, 6xy could become (6x)(y). Then use the product rule. Derivative of 6xy with respect to x would end up being 6y + 6x(dy/dx).

Here's a pretty simple example for implicit differentiation: finding the derivative of x^2 + y^2 = 16.

(d/dx)(x^2) + (d/dx)(y^2) = (d/dx)(16)
2x + 2y(dy/dx) = 0
(dy/dx)(2y) = -2x
(dy/dx) = (-2x)/(2y)
(dy/dx) = -(x/y)

I'm too lazy to find and type up a harder example... but hopefully you get the idea. You have to differentiate each term separately using the rules that I stated above (well, I kind of made them up as I went along so I hope they work all the time), and then rearrange the equation to get an equation for dy/dx in terms of x and y.

If you have any more questions, just ask!