Why does implicit differentiation work? That is why is ?
Does that mean that is a function but and are not?
Two questions Watch
- Thread Starter
- 11-11-2008 23:06
- 11-11-2008 23:14
Last edited by BJack; 11-11-2008 at 23:34.
- 11-11-2008 23:17
1. Consider the chain rule.
2. They are not functions (where the definition of function requires one-one mapping).
- 11-11-2008 23:18
1) It's a result of the chain rule. Write
z = f(y)
Now we're trying to find dz/dx, which we know is
dz/dx = dz/dy * dy/dx (chain rule)
But dz/dy = f'(y), so
dz/dx = f'(y) * dy/dx, as required.
2) Haven't read the article but y = x^2 is a many-to-one function (i.e. many values can lead to the same answer, e.g. (-5)^2 = 5^2 = 25). x = y^2 is not a function of y and neither is the other one.
If one value of x can lead to more than one value of y, then it's not a function.