# About the chain rule what's wrong with me?

1. Feb 16, 2012

### MHD93

What I know from the chain rule is that if y and u are differentiable with respect to x then dy/dx = (dy/du)*(du/dx)

Now, why is this example doesn't work:
y = x^2
u = c

then we have dy/dx = (dy/du) * (du/dx) = (dy/du) * 0 = 0 doesn't equal 2x
I want an answer irrelated to the chain rule, not explainations involving limits only.

THANKS

2. Feb 16, 2012

### chiro

Mohammad_93 and welcome to the forums.

For this particular problem, you will note that y is not defined in terms of u in any way that exhibits information about change: that is in terms of u against y, u does not contribute any information about y changes with respect to u or vice versa.

For this particular problem you can not use the chain rule for this reason.

An easy way to understand this is to consider a non-chain rule scenario y = c.

Think of the relationship dx/dy = 1/(dy/dx). We know that dy/dx = 0 because we have a flat function which means that we know how y changes with a specific change in x.

But we can't do the reverse. Given a change in y in this specific circumstance, we can not see how a change in y gives a subsequent change in x which is what the dx/dy is actually telling us.

Your chain rule problem is of a somewhat similar nature in that you are trying to find derivatives that do not make sense which means that you get a non-sensical answer.

You get the same kind of problems when we deal with division by 0. For this reason we do not define what dividing by zero actually is because it leads to all kinds of problems in a similar way to what you are getting with your calculus problem.

3. Feb 16, 2012

### Fredrik

Staff Emeritus
I prefer to write the chain rule in the form $(f\circ g)'(x)=f'(g(x))g'(x)$, where $f\circ g$ is the composition of f and g, defined by $f\circ g(x)=f(g(x))$ for all x. (The left-hand side is interpreted as $f\circ g$ acting on x). You wrote down two equalities that define two functions. If we call the first one f and the other one g, then $f\circ g$ would satisfy $f\circ g(x)=f(g(x))=f(c)=c^2$. The derivative of this $f\circ g$ is 0, and the chain rule confirms that: $(f\circ g)'(x)=f'(g(x))g'(x)=2c·0=0$. f'(g(x)) is =2c in this case, because for all t, f'(t)=2t, and if you insert t=g(x)=c, you get f'(g(x))=2c.

4. Feb 19, 2012

### noobilly

I suppose because u=c is not a function?

5. Feb 19, 2012

### willem2

u = c is a perfectly fine function of x.

It's because there is nu function f, such that f(u) = f(c) = x^2.

6. Feb 19, 2012

### alexfloo

"if y and u are differentiable with respect to x"

The issue is that dy/du does not exist. Look at the definition of the derivative:

lim (y(x) - y(x+h))/(u(x) - u(x+h)) = lim (y(x) - y(x+h))/(c - c),

so we are dividing by zero.

For the theorem to work, u must be nonconstant in some neighborhood of the point where we are taking the derivative. Otherwise we run into this issue above.