I understand that the derivative of a function can be evaluated from first principles. However I do not understand why dy/dt x dt/dx = dy/dx. Surely these terms cannot be treated in the same way in which you would multiply fractions. Could somebody please explain this intuitively?
60508
We know that the chain rule is used to differentiate  a composite function ,say $$f(x) = h(g(x))$$ It's defined as the derivative of the outside function times the derivative of the inner function or the other way around.  $$\frac{\mathrm{dy} }{\mathrm{d} x} = \frac{\mathrm{dy} }{\mathrm{d} u} \cdot \frac{\mathrm{du} }{\mathrm{d} x}$$  Despite we know that the above expression is not a fraction (even though it's a fractional notation of the derivative used by Leibnitz) you can "cancel" the two du's and get back dy/dx.  My question is: How can you even think of cancelling du from dy/du and du from du/dx when they are not even fractions. Just because it's been multiplied do they automatically become fractions?Are they really being "multiplied"?  I'am really looking for an intuition behind this.To me this is some kind of fantasy.It doesn't appear to be real.
89574
I've been having a few questions about infinite derivative for a long time now. I think to answer these questions involves dealing with properties of limit, hope someone can help. Thanks!  1). Does writing the derivative of a function as a function lose information? For example, the function $y = x^{1/3}$ has a derivative that evaluates to infinite at $x = 0$. But its derivative is undefined at the point, we only know it $\lim_{x\to0} y$ is infinite.   2). I know that most calculus textbooks consider having an infinite derivative to be not differentiable at that point, which means we are not allowed to apply chain rule and product rule. However, am I guaranteed to get a wrong answer if I apply them to cases where derivatives are infinite, or are there cases where it is still legal to do so(by the nature of dealing with infinite limits)?  3). The slope at $t = c$ on a curve defined by parametric equations $y = f(t)$ and $x = g(t)$ is $dy/dx = (dy/dt)/(dx/dt)$. When both $dy/dt$ and $dx/dt$ equals $0$ at $t = c$, the textbook applies l'Hospital's rule to find $\lim_{t\to c} dy/dx$ and thereby determines the slope at $t = c$. But why should we turn to find the limit of the derivative as $t\to c$ instead of the derivative itself at $t = c$?
71708
I ask because, as a first-year calculus student, I am running into the fact that I didn't quite get this down when understanding the derivative:  So, a derivative is the rate of change of a function with respect to changes in its variable, this much I get.  Thing is, definitions of 'differential' tend to be in the form of defining the derivative and calling the differential 'an infinitesimally small change in x', which is fine as far it goes, but then why bother even defining it formally outside of needing it for derivatives?  And THEN, the bloody differential starts showing up as a function in integrals, where it appears to be ignored part of the time, then functioning as a variable the rest.  Why do I say 'practical'? Because when I asked for an explanation from other mathematician parties, I got one involving the graph of the function and how, given a right-angle triangle, a derivative is one of the other angles, where the differential is the line opposite the angle.   I'm sure that explanation is correct as far it goes, but it doesn't tell me what the differential DOES, or why it's useful, which are the two facts I need in order to really understand it.  Any assistance?
8534
What is the difference between implicit, explicit, and total time dependence, e.g. $\frac{\partial \rho}{\partial t}$ and $\frac{d \rho} {dt}$?  I know one is a partial derivative and the other is a total derivative. But physically I cannot distinguish between them. I have a clue that my doubt really might be understanding the difference between implicit, explicit and  total time dependence.
42960
One problem I have found when learning calculus is that there are many different ways to denote the derivative. If $y=f(x)=x^2$, then we could write \begin{align} f'(x)&amp;=2x \\ y'&amp;=2x \\ \frac{df}{dx}(x)&amp;=2x \\ \frac{df(x)}{dx}&amp;=2x \\ \frac{d}{dx}f(x)&amp;=2x \\ \frac{dy}{dx}&amp;=2x \end{align} And this is just Lagrange and Leibniz's notations alone. What I find troubling is that they all seem to be suggesting subtly different things about what the derivative actually is. Is it a function, a limit of a quotient, or both? In the interests of keeping my post brief, I'll focus my attention on $f'(x)=2x$ and $\frac{dy}{dx}=2x$, as these seem to be the most common notations.  $$ f'(x)=2x $$  It does make sense to think of the derivative as the gradient function: $$ f'\colon x\mapsto\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} $$ In this case the limit expression is equal to $2x$, and so we can write $$ f' \colon x \mapsto 2x $$ However, this notation seems a little counter-intuitive when we consider what it means to differentiate a function with respect to a variable other than $x$. If I ask what is the derivative of $f(x)$ with respect to $\frac{x}{2}$, does this question make sense? Is it simply $f'(\frac{x}{2})$? Or do we have to express $x^2$ in terms of $\frac{x}{2}$? And how can we can express this derivative using Lagrange's notation?  $$ \frac{dy}{dx}=2x $$  There are many things which are nice about Leibniz's notation, including the fact that it is explicit which variable you are differentiating with respect to. However, in this case, it is unclear whether we are talking about a function, or something else entirely. There are other issues. Some people say they dislike the Leibniz formulation of the chain rule $$ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} $$ saying that they find it to be inaccurate. I don't really understand why this is the case. Could someone please elaborate?
3973
I do not know much about fractional calculus, except what I have read in a few short posts at MSE and .  I know that order of a derivative can be extended from rational values to real values, but all I know is what is written .  So my question is, what is the simplest way to understand and/or define $\dfrac{d^n}{dx^n}f(x),$ for $n\in\mathbb{R}$?  For example what is $\dfrac{d^\pi}{dx^\pi}f(x)$?  Also, what about complex values?  What might we say about $\dfrac{d^{s}}{dz^s}f(z)$ for $s\in\mathbb{C}$?  And what are some uses for such derivatives?
27764
In lectures we're told:  $$\dfrac {\partial y} {\partial x} = \dfrac 1 {\dfrac {\partial x} {\partial y}}$$  as long as the same variables are being held constant in each partial derivative.  The course is 'applied maths', i.e non-rigorous, so don't confuse me.  But anyway, if we have:  $$\xi = x - y \qquad \eta = x$$  Then $\dfrac {\partial x} {\partial \xi} = 0$ and $\dfrac {\partial \xi} {\partial x} = 1$.  The rule presumably fails because one of the partial derivatives is $0$.  But then isn't this rule useless? Since I cannot use it without checking that the partial derivative isn't in fact $0$, but then I've worked out the partial derivative manually anyway.
113196
Okay, perhaps not a paradox, but somewhat of a lack of understanding on my part.  Let $z$ equal some function of $x$ and $y$, i.e. $z = f(x, y)$ and take partial derivatives $\frac{\partial z}{\partial x} = f_x$ and $\frac{\partial z}{\partial y} = f_y$ all and good. But now say I do partial differentiation with respect to z. $\frac{\partial z}{\partial z} = f_z$ which equals $0$ because $f(x, y)$ is a function of $x$ and $y$, but not $z$, so all $x$ and $y$ are held constant and the derivative of a constant is zero. But that's not the right answer is it? $\frac{\partial z}{\partial z}$ should be equal to 1. Where did my logic go wrong?
113392
Given the following equation:  $f(xy)=f(x)+f(y)$  and the fact that $y=x^{-1}$,  I've to find how this could became:  $f'(x) = f'(1)/x$,  where it is said that $f'(x)$ is the total derivative of $f$  I really miss some steps..someone could help?
106400
How would I find the derivative of a unit step function? I understand that the unit impulse function will be used but I'm not sure how to use it.  I am trying to find the derivative of this:  $v(t) = u(t+1) - 2u(t) + u(t-1)$  $u(t) = 0$  when $t &lt; 0$  $u(t) = 1$ when $t &gt; 0$  The relationship between unit step function and impulse function:  &delta;(n) = u(n) - u(n-1)  $ δ(t)=du(t)/dt $
94839
A friend told me to solve the following differential equation:  $$f'(x)=f(f(x))$$  I have no idea how to solve this! This doesn't seem to be an ordinary differential equation and I can't even solve this numerically!  I think my friend is trolling me.
121843
$$\frac{dx(t)}{dy(t)}=\frac{\alpha x(t) - \beta x(t) y(t)}{-\gamma y(t) + \delta x(t)y(t)}$$  How would one simplify this fraction? Maybe the chain rule could be of any use, but I don't see how.
75597
I am having trouble finding the derivative of the following:  $$y = x^x+x^3+3^x+3^3$$  $$\frac{dy}{dx}= x \times x^{(x-1)}+3x^2+3^x\ln3+0$$  I think the $x^x$ part is wrong. Any help would be appreciated.
58372
I've been trying to understand how the second order derivative "formula" works:  $$\lim_{h\to0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}$$  So, the rate of change of the rate of change for an arbitrary continuous function. It basically feels right, since it samples "the after $x+h$ and the before $x-h$" and the $h^2$ is there (due to the expected /h/h -> /h*h), but I'm having trouble finding the equation on my own.  It's is basically a derivative of a derivative, right? Newtonian notation declares as $f''$ and Leibniz's as $\frac{\partial^2{y}}{\partial{x}^2}$ which dissolves into:  $$(f')'$$ and $$\frac{\partial{}}{\partial{x}}\frac{\partial{f}}{\partial{x}}$$  So, first derivation shows the rate of change of a function's value relative to input. The second derivative shows the rate of change of the actual rate of change, suggesting information relating to how frequenly it changes.  The original one is rather straightforward:  $$\frac{\Delta y}{\Delta x} = \lim_{h\to0} \frac{f(x+h) - f(x)}{x + h - x} = \lim_{h\to0} \frac{f(x+h) - f(x)}{h}$$  And can easily be shown that $f'(x) = nx^{n-1} + \dots$ is correct for the more forthcoming of polynomial functions. So, my logic suggests that to get the derivative of a derivative, one only needs to send the derivative function as input to finding the new derivative. I'll drop the $\lim_{h\to0}$ for simplicity:  $$f'(x) = \frac{f(x+h) - f(x)}{h}$$  So, the derivative of the derivative should be:  $$f''(x) = \lim_{h\to0} \frac{f'(x+h) - f'(x)}{h}$$  $$f''(x) = \lim_{h\to0} \frac{  \frac{ f(x+2h) - f(x+h)}{h} - \frac{ f(x+h) - f(x)}{h}  }{h}$$  $$f''(x) = \lim_{h\to0} \frac{  \frac{ f(x+2h) - f(x+h) - f(x+h) + f(x)}{h}  }{h}$$  $$f''(x) = \lim_{h\to0} \frac{  f(x+2h) - f(x+h) - f(x+h) + f(x)  }{h^2}$$  $$f''(x) = \lim_{h\to0} \frac{  f(x+2h) - 2f(x+h) + f(x)  }{h^2}$$  What am I doing wrong? Perhaps it is the mess of it all, but I just can't see it. Please help.
118612
Is there a simple and intuitive way to prove the chain rule, that is, if $y$ is a function of $u$ and $u$ is a function of $x$, then why is $\frac{dy}{dx}$ = $\frac{dy}{du}$ $\cdot$ $\frac{du}{dx}$ ? This could just be an intuitive argument.   PS: The only proofs I found were based off of confusing definitions.
42686
In elementary calculus (and often in courses beyond) we are taught that a differential of a function, $df$ quantifies an infinitesimal change in that function. However, the notion of an infinitesimal is not well-defined and is nonsensical (I mean, one cannot define it in terms of a limit, and it seems nonsensical to have a number that is smaller than any other real number - this simply doesn't exist in standard analysis). Clearly the definition $$df=\lim_{\Delta x\rightarrow 0}\Delta f =f'(x)dx$$ makes no sense, since, in the case where $f(x)=x$ we have that $$ dx=\lim_{\Delta x\rightarrow 0}\Delta x =0.$$  All of this leaves me confused on how to interpret expressions such as $$df=f'(x)dx$$ Should it be seen simply as a definition, quantifying the first-order (linear) change in a function about a point $x$? i.e. a new function that is dependent both on $x$ and a finite change in $x$, $\Delta x$, $$df(x,\Delta x):=f'(x)\Delta x$$ then one can interpret $dx$ as $$dx:=dx(x,\Delta x)= \Delta x$$ such that $$\Delta f=f'(x)dx+\varepsilon =df +\varepsilon$$ (in which $\varepsilon$ quantifies the error between this linear change in $f$ and the actual change in $f$, with $\lim_{\Delta x\rightarrow 0}\varepsilon =0$).  I feel that there must be some sort of rigorous treatment of the notion of differentials since these kind of manipulations are used all the time, at least in physics?!  I've had some exposure to differential geometry in which one has differential forms, in particular $1-$forms which suggestively notationally "look like" differentials, for example $$\omega =df$$ but as I understand it these are defined as linear maps, members of a dual space to some vector space, $V$, which act on elements of $V$, mapping them to real numbers. Furthermore, the basis $1-$forms are suggestively written as what in elementary calculus one would interpret as an infinitesimal change in x, $dx$. But again, this is simply symbolic notation, since the basis $1-$forms simply span the dual space and are themselves linear maps which act on elements of $V$.  I've heard people say that differential forms make the notion of a differential of a function mathematically rigorous, however, in my mind I can't seem to reconcile how this is the case, since at best they specify the direction in which the differential change in a function occurs, via $$df(v)=v(f)$$ (since $v(f)$ is the directional derivative of a function $f$ along the vector $v$).  If someone could enlighten me on this subject I'd really appreciate it.
1036
I want to differentiate   $x^2 + y^2=1$ with respect to $x$.  The answer is $2x +2yy' = 0$.  Can some explain what is implicit differentiation and from where did $y'$ appear ? I can understand that $2x +2yy' = 0$ is a partial derivative but then it becomes multi calc not single.  This is in a chapter about chain rule so I assume there is some use of the chain rule here but I can't see any composite functions here. We can express y in terms of x but y is not composite.  P.S: I am NOT looking for HOW to solve the problem, I am looking for the WHY as stated above.