Question

My own thought regarding implicit differentiation: when you get the derivative of y^2=0, you get 2y * y' =0.

So if you find the derivative of just y. Does it equal to y' or y*y'? Please explain why as well.Thank you!

Answer 1

its just y'
think of it this way
y=5x
dy/dx = 5
you differentiated "implicitly" the y on the left hand side

Answer 2

You need to satisfy the requirements for a derivative before you start "taking" derivatives. y^2 = 0 defines only a single value, y = 0. This is not differentiable.

Answer 3

I still don't understand "y=5x
dy/dx = 5
you differentiated "implicitly" the y on the left hand side"

@tkhunny: I don't know, this was in my class notes :O

Answer 4

You have confused the power rule. \(\frac{dy}{dx}=y′\) but \(\frac{d(y^2)}{dx}=2yy′\)

Answer 5

Can I think of it as:
if y=5x
and y'=5

So
y^1 = 1y^0 * y'
Like how:
y^2 = 2y^1 * y'?

Answer 6

Start with a continuous function, which, at least locally, is differentiable.

Shadowys has brilliantly avoided entirely any such concern, but how? Simply avoiding the odd "=" was sufficient. We're just talking about some function and not an equation that cannot be construed as a function.

Answer 7

Yes. That's the power rule. you've just rederived it.

Answer 8

So wait, what does Shadowys mean when he says " You have confused the power rule.. dydx=y' but d(y2)dx=2yy'" I think the "but" is throwing me off.

Answer 9

oh wait Hunny, if anything my notes say
y^2 = 0
2y*y' = 0
y' = 0

And so it is not differentiable because y' is like a slope. And when a slope is 0, it is horizontal....

Can you define "differentiable"? I never understood the term even though i learned it last month >.<

Answer 10

It means that differentiating some function that is a variable of another function doesn't just involve simple \(nx^{n-1}\) In this case, I'm assuming that y in a function of another variable, maybe x.
i.e.
\(y=f(x)=y \) and
\(u=g(y)=y^2\)
when differentiating wrt x, both of them involve different ways. One is just applying formula, the other needs the power rule.

Answer 11

I'm sorry I'm not sure if I understand your previous comment (fully anyway). But can i assume along the lines of (Because i kept staring at your "confuse the power rule" problem")
So dy/dx=y' Means the derivative of y
and d(y2)/dx=2yy' Means the derivative of y^2 plus the chain rule (or was it power rule...)
?

Answer 12

yup! :) that means \(\frac{dy}{dx}\) will never be yy'

Answer 13

Omg thank you. It's funny how a small question helps me solve my other problems in calc as well...

Answer 14

Implicit Differentiation is a study in the Chain Rule. The power rule is only a coincidence because you started with y^2. Had you started with ln(y), you would not have used the power rule.

Answer 15

lol Yeah maths' like that. Good luck