Differentiation question

[cmkessell]

My son just asked me for help with a differentiation question (it's not a sneaky homework question, just part of his revision for forthcoming exams). It's been 20+ years since I did calculus so I had to give up.

Basically the question is to find dy/dx in terms of x and y for the equation:

y^2 + y = x^3 + 2x

I'm sure it's straightforward but just can't recall the technique to use. Some clues on how to solve it would be appreciated.

[jemidiah]

Technique: chain rule and product rule.

Application:
Recall for any variable y we have a formula that gives the derivative of y^n with respect to y as n*y^(n-1). It turns out for taking derivatives of an expression containing some variable with respect to a different variable, an extra term arises--here, dy/dx.

So from the chain and product rules, d/dx (y^2) = 2y*dy/dx = 2y y'. Then the rest is straightforward:

y^2+y = x^3+2x
2y y' + y' = 3x^2 + 2
y'(1+2y) = 3x^2+2
y' = (3x^2+2)/(1+2y)

Sometimes the chain rule is put as df/dy*dy/dx = df/dx for making it easy to remember. (In my opinion the notation is unfortunate because it causes people to ignore the proof, and maybe think that differential elements behave like fractions, but I'm in the vast minority I'm sure.) Here, d(y^2)/dx = d(y^2)/dy*dy/dx = 2y y'.

[NickThissen]

This is called implicit differentiation. It's really not that hard, as long as you are aware what you are doing each time.

You need to differentiate both sides (each term) to x. The right side is easy and is just normal differentiation:
d/dx ( x^3 + 2x ) = 3x^2 + 2

The left side does not contain x explicitly, but it does contain x implicitly, because y is a function of x!
So differentiating the left side to x requires the chain rule.

First the derivative of y^2. Remember that y is a function of x: y = y(x).

If you differentiate a function of x, you just need to apply the chain rule. The chain rule is usually shown as follows:
d/dx ( f(g(x)) ) = df/dg * dg/dx

For example, if f(x) = 1 + 3x^2, and g(x) = 4x + 2, then
f(g(x)) = f(4x + 2) = 1 + 3(4x + 2)^2
or: f(g) = 1 + 3g^2

The derivative of the last f(g(x)) = 1 + 3(4x + 2)^2 is d/dx(1 + 3(4x+2)^2) = 6(4x+2)*4 = 24(4x + 2)
The derivative of f(g(x)) = df/dg dg/dx = 6g 4 = 6(4x+2)*4 = 24(4x + 2).
So they are equal.


Now, apply the same to y, regarding y as a function of x: y = y(x).
Let's call y^2 a different name, let's say f(x), for example.

d/dx ( y^2 ) = d/dx ( f ) = df/dx = ... chain rule ... = df/dy dy/x
df/dy is just 2y (note that you are now differentiating to y!) so :
d/dx ( y^2 ) = 2y dy/dx.


The second term on the left hand side is just y, and its derivative is just dy/dx.

So we have after differentiating both sides:

2y dy/dx + dy/dx = 3x^2 + 2

You can solve this for dy/dx:
dy/dx(2y + 1) = 3x^2 + 2
dy/dx = (3x^2 + 2) / (2y + 1)

And there you have expressed dy/dx in terms of x and y.


Another way to do this that would have to give the same results is by expressing y as a function of x explicitly. y(x) = f(x).
This may not always be possible however, for example the function
log(y) + sin(y) + y^2 = 3x + e^x

It is not possible to express y as an explicit function of x here, so we need to use implicit differentiation.

[jemidiah]

log(y) + sin(y) + y^2 = 3x + e^x

It is not possible to express y as an explicit function of x here, so we need to use implicit differentiation.

In the technical sense, one could express y as the limit of some infinite series of elementary functions and get an explicit (but remarkably useless) expression for y as a function of x. It would require tons more theorems and exploration into derivatives, though, so it's simply not done. Good writeup

[cmkessell]

Many thanks for the responses. I understand it now, and I'll ensure my son brushes up on his implicit differentiation and chain/product rules before his exam!