# A derivative question

#### mech-eng

First I am sorry if this quesiton is futile, or even stupid but I find it confusing, and I would like to clarify in my mind.

Taking the derivative $y=x^2$ with respect to x by using limit is very easy stuff. But would you please illustrate how to apply this to the equation $y^2=x^2$

It seems there is no plate anymore. For an attempt I can just apply it to the simple equation $y=x^2$.

$(x+h)^2-x^2/h$ as $h$ goes to zero. $(x^2+2xh+h^2 - x^2)/h$ as $h$ goes to zero. The two $x^2$ cancels each other. $(2xh+h^2)/h$ as $h$ goes to zero. Cancelling $h$ on the denominatior and numerator, $2x$.

Thank you.

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#### RPinPA

Are you talking about deriving implicit differentiation using the definition of derivative? Let's first take a look at how implicit differentiation is usually done.

You have the equation $y^2 = x^2$ which defines $y$ implicitly by its relation to $x$, and you'd like to know $\frac {dy}{dx}$. So we differentiate both sides.

On the left we have $\frac {d(y^2)} {dy} \frac {dy} {dx}$ by the chain rule, and on the right we have $2x$. So we get $2y \frac {dy} {dx} = 2x$.

If that's what you're trying to obtain from the definition of derivative, then you're going to have to derive the chain rule along the way. Is that what you're trying to do?

#### mech-eng

Are you talking about deriving implicit differentiation using the definition of derivative? Let's first take a look at how implicit differentiation is usually done.

You have the equation $y^2 = x^2$ which defines $y$ implicitly by its relation to $x$, and you'd like to know $\frac {dy}{dx}$. So we differentiate both sides.

On the left we have $\frac {d(y^2)} {dy} \frac {dy} {dx}$ by the chain rule, and on the right we have $2x$. So we get $2y \frac {dy} {dx} = 2x$.

If that's what you're trying to obtain from the definition of derivative, then you're going to have to derive the chain rule along the way. Is that what you're trying to do?

Yes, but here I cannot see the use of limit. Thank you for introducing chain rule here. It made the things more clear to me now. But how can we take that derivative by giving an increment"h" to the variables?

#### fresh_42

Mentor
2018 Award
Taking the derivative $y=x^2$ with respect to $x$ by using limit is very easy stuff. But would you please illustrate how to apply this to the equation $y^2=x^2$.
This is not a function anymore. However, since derivatives are local calculations, we can solve the equation locally and get $y=x$ or $y=-x$. Now you can apply the limit definition, except at $x=0$ where the limits are not defined because the way to approach it isn't unique.

#### mech-eng

This is not a function anymore.
Since this is not a "function", do you mean this is a "relation"? Then we can locally treat relations as functions, right?

Thank you.

#### fresh_42

Mentor
2018 Award
Since this is not a "function", do you mean this is a "relation"? Then we can locally treat relations as functions, right?

Thank you.
Not all relations. E.g. at $x=0$ we do not get a local function, it remains a relation. The graph is an X and we cannot get rid of the ambiguity at the intersection. Relations can be quite wild, so I wouldn't say that it's always possible. But the essential point is the locality. Derivatives are at a certain point, that is a local property. That we write $y'=2x$ is a bit sloppy, or an abbreviation for the function $x \longmapsto \left. \dfrac{d}{dt} \right|_{t=x}y(t)$ if you like this better.

#### Ray Vickson

Homework Helper
Dearly Missed
First I am sorry if this quesiton is futile, or even stupid but I find it confusing, and I would like to clarify in my mind.

Taking the derivative $y=x^2$ with respect to x by using limit is very easy stuff. But would you please illustrate how to apply this to the equation $y^2=x^2$

It seems there is no plate anymore. For an attempt I can just apply it to the simple equation $y=x^2$.

$(x+h)^2-x^2/h$ as $h$ goes to zero. $(x^2+2xh+h^2 - x^2)/h$ as $h$ goes to zero. The two $x^2$ cancels each other. $(2xh+h^2)/h$ as $h$ goes to zero. Cancelling $h$ on the denominatior and numerator, $2x$.

Thank you.
Since your example really says that either $y=x$ or $y=-x,$ let us make it a bit more involved by taking, instead, the example $y^2 = x^3.$ We have
$$(y + \Delta y)^2 - y^2 = (x + \Delta x)^3 - x^3,$$
which expands out to
$$2 y \Delta y + (\Delta y)^2 = 3 x^2 \Delta x + 3 x (\Delta x)^2 + (\Delta x)^3.$$
Let us keep only the first-order increments:
$$2 y \Delta y \doteq 3 x^2 \Delta x,$$
which will become exact in the limit. Thus, we have
$$2 y \frac{\Delta y}{\Delta x} \doteq 3 x^2,$$
so in the limit as $\Delta x \to 0$ we get
$$2 y \frac{dy}{dx} = 3 x^2 \; \Leftarrow\: \text{exact!}$$

You would get this immediately from the chain rule:
$$\frac{dy^2}{dx} = \frac{d x^3}{dx},$$ or
$$\underbrace{\frac{dy^2}{dy}}_{2 y} \cdot \frac{dy}{dx} = 3 x^2$$

#### mech-eng

Since your example really says that either y=xy=xy=x or y=−x,y=−x,y=-x, let us make it a bit more involved by taking, instead, the example y2=x3.y2=x3.y^2 = x^3. We have
(y+Δy)2−y2=(x+Δx)3−x3,​
Thanks for the explanation but would you please expand on how to know the differences on the right and left is equal? That is how can we be sure
(y+Δy)2−y2=(x+Δx)3−x3,​

#### Ray Vickson

Homework Helper
Dearly Missed
Thanks for the explanation but would you please expand on how to know the differences on the right and left is equal? That is how can we be sure

(y+Δy)2−y2=(x+Δx)3−x3,​
Because I want both points $(x,y)$ and $(x + \Delta x, y + \Delta y)$ to be on the curve $y^2 = x^3$.

#### mech-eng

Since your example really says that either y=xy=xy=x or y=−x,y=−x,y=-x, let us make it a bit more involved by taking, instead, the example y2=x3.y2=x3.y^2 = x^3. We have
(y+Δy)2−y2=(x+Δx)3−x3,(y+Δy)2−y2=(x+Δx)3−x3,​
(y + \Delta y)^2 - y^2 = (x + \Delta x)^3 - x^3,
which expands out to
2yΔy+(Δy)2=3x2Δx+3x(Δx)2+(Δx)3.2yΔy+(Δy)2=3x2Δx+3x(Δx)2+(Δx)3.​
2 y \Delta y + (\Delta y)^2 = 3 x^2 \Delta x + 3 x (\Delta x)^2 + (\Delta x)^3.
Let us keep only the first-order increments:
2yΔy≐3x2Δx,​
Would you please explain why we prefer only first-order increments? What if we do not neglect the second-order ones?

Thank you.

#### fresh_42

Mentor
2018 Award
Would you please explain why we prefer only first-order increments? What if we do not neglect the second-order ones?
The entire concept is a linear one. We try to find a linear equation which locally approximates the given function. This is because linear functions are easier to handle and in a local neighborhood of the point we differentiate at they are good enough. There is no second order term.

The expansion of a function into its Taylor series is another subject, it's an iteration of approximations, a refinement of the process. In physics, second or higher order terms (of named Taylor expansion) can be useful, but this has nothing to do with the process of differentiation.

#### mech-eng

The entire concept is a linear one. We try to find a linear equation which locally approximates the given function. This is because linear functions are easier to handle and in a local neighborhood of the point we differentiate at they are good enough. There is no second order term
By entire concept is a linear one, do you mean derivatives are the slopes of tangent lines and the tanget lines are linear, and tangent lines are good approximation to a curve immediately around a point at which they touch?

#### fresh_42

Mentor
2018 Award
By entire concept is a linear one, do you mean derivatives are the slopes of tangent lines and the tanget lines are linear, and tangent lines are good approximation to a curve immediately around a point at which they touch?
Yes. That is one way to look at it. The limit definition you started with is that of a family of secants which end up in a tangent. You can write the defining equation as $\mathbf{f(x_{0}+v)=f(x_{0})+J(v)+r(v)}$ where $J$ is the linear function, the derivative, $v$ the direction in which we differentiate, and $r$ the error we make by this approximation.

For the example $f(x)=y=x^2$ we get
$$\underbrace{(x_0+h)^2}_{=f(x_0+v)} = \underbrace{x_0^2}_{=f(x_0)} + \underbrace{2x_0\cdot }_{= J= \text{ times }2x_0} \underbrace{h}_{=v} + \underbrace{r(v)}_{=h^2}$$
We approximated $f(x)$ at $x=x_0$ with the linear function $J$, which is times $\mathbf{2x_0}$, i.e. $h \longmapsto 2x_0\cdot h$ and an error $h^2$.

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