What is implicit differentiation?

1 Answer
Jan 1, 2016

I tried this hoping it is understandable!

Explanation:

You may remember the difference between an implicit and explicit form of a function:
Normally your function can be read explicitly as
#y="something of x"#
and derived immediately to find #(dy)/(dx)="something else of x"#;

For example #y=x^2-3x#
#(dy)/(dx)=2x-3#

Sometimes your function is more difficult to "extricate" and can be difficult to write as #y="something of x"#.
You can only write it as #y(x)+ax^n+...c=0#
As you can see the #y# is nested inside the other bits and it is in itself a function of #x#.

Consider for example a circle:
#x^2+y^2=4#
Here #y# is "difficult" to extract (ok, not impossible) and if you try to derive as it is, you have to remember to derive also #y#:
#2x+color(red)(2y(dy)/(dx))=0#
Rearranging:
#(dy)/(dx)=-(2x)/(2y)=-x/y#
As you can see this derivative contains again #y#; once you can figure out the form of #y# you could substitute it in and get the usual derivative containing only #x#.

In the example of the circle you can write it as (extracting #y#):
#y=+-sqrt(4-x^2)#
So that gives you:
#(dy)/(dx)=-x/(color(red)(+-sqrt(4-x^2))#

It isn't always possible to extract #y# and you leave the derivative as it is, with #y# in it.

To test this result try to derive the explicit #y=+-sqrt(4-x^2)# form and see if it checks with the one we found implicitly.