The chain rule involves two variables, one independent and one dependent on the independent one. Normally these are called x (independent) and y (dependent).
If y = sin(4x) we can't do anything straightforward so we make up an intermediate variable u and define u as u = 4x.
The chain rule states that dy/dx = dy/du * du/dx
------------ Chain Rule
(almost as if you were multiplying fraction although **that is not what we're doing**.
So u = 4x ........................and du/dx = 4
Now we have the simple y = sin(u), so......................dy/du
So dy/dx = dy/du * du/dx = sin(u) * 4 = 4 * sin(u). But u was an intermediate variable and u = 4x, so where we have sin(u), we now have sin(4x).
That means that dy/dx = 4 * sin(u) is really dy/dx = 4 * sin(4x)
---- End of chain rule.
---------------------------------
However we may run into something where both x and y have complicated places, like:
((x^2)*(y^3)) - (x*sin(y)) = 78.............Find dy/dx
The chain rule cannot handle this so we use implicit differentiation as well as the product rule.
First let's differentiate ((x^2)*(y^3))
The product rule says that this is (y^3 * 2x) + ((x^2 * "what"))
- Critical
That "what" is where implicit differentiation comes in. Since x is the independent variable the derivative of y^3 is 3y^2 * dy/dx (the dy/dx is necessary because it is x that is independent.)
So "what" is 3y^2 dy/dx and looking back at
--- Critical we see that becomes:
(y^3 * 2x) + ((x^2 * 3y^2 * dy/dx))
That takes care of the ((x^2) * (y^3)), but what about the other term:
- (x*sin(y)). Again the product rule rears its ugly head.
-(sin(y) + (x*cos(y) * dy/dx)
The derivative of sin(y) is cos(y) but again we need to put in the dy/dx because it is a y term that is being differentiated.
Of course the derivative of 78 is 0, since 78 is a constant.
So putting this altogether we have:
(y^3 * 2x) + ((x^2 * 3y^2 * dy/dx)) - (sin(y) + (x*cos(y) * dy/dx) = 0
From here we want to isolate the dy/dx so put the dy/dx terms on the LHS and the non-dy/dx on the RHS. We do this through adding and subtracting.
[((x^2 * 3y^2 * dy/dx) + (x * cos(y) * dy/dx)] = [sin(y) - 2xy^3]
I'm going to make x^2 * 3y^2 * dy/dx = 3x^2 * y^2 * dy/dx
Now let's factor the LHS
dy/dx * [3x^2y^2 + x*cos(y)] = [sin(y) - 2xy^3] and with a quick division.
dy/dx = [sin(y) - 2xy^3] / [3x^2y^2 + x*cos(y)]
--- Answer
As you can see implicit differentiation is much more complicated than using the chain rule. It should be, it's reserved for problems where you have complicated terms of both x and y, while the chain rule has y by itself.
You can also see how many more steps the implicit differentiation took than the application of the chain rule.
Sadly enough this really was a simple problem for implicit differentiation. Remember, you will be using the product rule a lot in implicit differentiation.
It's tough, but these things happen, even on exams. Practice some of them and be ready before your next exam.
Good Luck,
Gerry
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