I thought I knew that we use differentials just because they made calculus intuitive and there was math rigor behind all of these. But when I saw   this problem came out.  For example, deriving implicitly: $$ x^2+y^2 = 5  $$ we get $$ 2x\mathrm{d}x +2y\mathrm{d}y = 0 $$ $$ 2xdx+2y\frac{dy}{dx}dx=0 $$ and then to find the slope of the graph we solve for dy/dx $$ \frac{dy}{dx}=-\frac{x}{y} $$  Does't that imply that $dx \neq 0$ ?  And how would it be in the general case whe differentiating implicitly
7365
In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $dy/dx$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $dy = f'(x)dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential). Then if we rearrange we get $dy/dx$ which could be seen as a ratio.   I wonder if the author says this because $dx$ is an independent variable, and $dy$ is a dependent variable, for $dy/dx$ to be a ratio both variables need to be independent.. maybe?
24671
Can any one explain me what is the use of derivatives in real life. When and where we use derivative, i know it can be used to find rate of change but why?. My logic was in  real life most of the things we do are not linear functions and derivatives helps to make a real life functions into linear.  eg converting parabola into a linear function $x^{2}\rightarrow 2x$  but then i find this; derivation of $\sin{x}\rightarrow \cos{x}$  why we cant use $\sin{x}$ itself to solve the equation. whats the purpose of using its derivative $\cos{x}$. Please forgive me if i have asked a stupid questions, i want to improve my fundamentals in calculus.
54292
In my Calculus class, my math teacher said that differentials such as $dx$ are not numbers, and should not be treated as such.  In my physics class, it seems like we treat differentials exactly like numbers, and my physics teacher even said that they are in essence very small numbers.   Can someone give me an explanation which satisfies both classes, or do I just have to accept that the differentials are treated differently in different courses?  P.S. I took Calculus 2 so please try to keep the answers around that level.  P.S.S. Feel free to edit the tags if you think it is appropriate.
101867
I can't get any of these on my own and I am attempting to do $x^x$ I had it explained to me for about an hour and I still can't do it on my own. I thought I was supposed to make it into $x(\ln x)$ which should be equivalent to the original term. Then the differentiation should be easy for most people from here.  $$1(\ln x)+x(\frac1x) $$ or $$\ln x+1$$  This is of course wrong but I do not know why.
29673
I was messing around with the definition of the derivative, trying to work out the formulas for the common functions using limits. I hit a roadblock, however, while trying to find the derivative of $e^x$. The process went something like this:  $$\begin{align} (e^x)' &amp;= \lim_{h \to 0} \frac{e^{x+h}-e^x}{h} \\ &amp;= \lim_{h \to 0} \frac{e^xe^h-e^x}{h} \\ &amp;= \lim_{h \to 0} e^x\frac{e^{h}-1}{h} \\ &amp;= e^x \lim_{h \to 0}\frac{e^h-1}{h} \end{align} $$  I can show that $\lim_{h\to 0} \frac{e^h-1}{h} = 1$ using L'Hôpital's, but it kind of defeats the purpose of working out the derivative, so I want to prove it in some other way. I've been trying, but I can't work anything out. Could someone give a hint?
24556
I am using two books for my calculus refresher.   Thomas' Calculus  Higher Math for Beginners by Ya. B. Zeldovich   My question is : When applying Integral Calculus for calculation of volumes of solids, generated by curves revolved around an axis, we use slices of 'cylinders' to approximate the volume of the resulting solid and then integrate the sum of those infinitesimal cylinders. However, when we are using the same techniques to calculate the surface area of the surfaces generated by revolving curves around an axis, we consider the 'slope' of the differential length of curve 'dx', calculate the total length of the curve and derive the required surface area.  Are we not supposed to use the same 'slope' for calculating the volumes of the infinitesimal 'cylinders' for calculation of volumes? Shouldn't we use 'sliced portions of 'cones' as the infinitesimal volumes?? When it come to calculation of volumes of solids of revolution, why are we neglecting the slope of the curve for the differential length and simply assume that it is an infinitesimal cylinder??  Ex: Let us say we want to calculate the surface area and the volume of the solid generated when the parabola y = 10 . x^2 is revolved about the y-axis, with limits of x from 0 to 5.  In such cases, when calculating the volume of the solid, we consider infinitesimal 'cylinders', ignoring the 'slope' of the curve for the differential element 'dx', but when calculating the surface area, we consider the 'slope' of the differential element 'dx'.
72361
I am reading in a book about differentiating, but I am confused with one of the steps he takes.  We start with:  $$ \begin{align} y &amp;= x^{2} \\ y + \mathrm{d}y &amp;= (x + \mathrm{d}x)^2 \\ y + \mathrm{d}y &amp;= x^2 + x\mathrm{d}x + x\mathrm{d}x + (\mathrm{d}x^2) \end{align} $$  Now the author simplifies this to:  $$y + dy = x^2 + 2x\mathrm{d}x + (\mathrm{d}x^2)$$  I dislike how the middle term is simplified to $2x\mathrm{d}x$ instead of $2(x\mathrm{d}x)$, as I feel like it is more intuitive on what is going.  As in, $2$ of the term $x\mathrm{d}x$, instead of $2x\mathrm{d}x$.  But I fear writing it as $2(x\mathrm{d}x)$ may result in an incorrect distributive property.  Next, he omits the $(dx^2)$:  $$y + \mathrm{d}y = x^2 + 2x \mathrm{d}x$$  Subtract the original $y = x^2$:  $$\mathrm{d}y = 2x \mathrm{d}x$$  Now here is where I get confused:  $$\frac{\mathrm{d}y}{\mathrm{d}x} = 2x$$  How can he just divide both sides by $\mathrm{d}x$!?  If the original term was $2$ of $x\mathrm{d}x$, wouldn't it have to be written out as $2x * 2\mathrm{d}x$, and thus divide both sides by $2\mathrm{d}x$ instead?  I think the root of my confusion is how to properly simplify: $$x\mathrm{d}x + x\mathrm{d}x$$  I trust that he is right, but I am looking for an explanation of why his simplification can work, and why $2(x\mathrm{d}x)$ would be incorrect!  Thank you!
19934
Integration is supposed to be the inverse of differentiation, but the integral of the derivative is not equal to the derivative of the integral:     $$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\int f(x)\mathrm{d}x\right) = f(x) \neq \int\left(\dfrac{\mathrm{d}}{\mathrm{d}x}f(x)\right)\mathrm{d}x$$   For instance: $$\begin{align*} &amp;\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\int 2x+1\;\mathrm{d}x\right) &amp;&amp;= \dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^2+x+C\right) &amp;= 2x+1\\ &amp;\int\left(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(2x+1\right)\right)\mathrm{d}x &amp;&amp;= \int 2\;\mathrm{d}x &amp;= 2x+C\end{align*}$$  Why isn't it defined such that $\dfrac{\mathrm{d}}{\mathrm{d}x}a = \dfrac{\mathrm{d}a}{\mathrm{d}x}$, where $a$ is a constant, and $\int f(x)\;\mathrm{d}x = F(x)$? Then we would have: $$\begin{align*} &amp;\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\int 2x+1\;\mathrm{d}x\right) &amp;&amp;= \dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^2+x\right) &amp;= 2x+1\\ &amp;\int\left(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(2x+1\right)\right)\mathrm{d}x &amp;&amp;= \int \left(2+\dfrac{\mathrm{d1}}{\mathrm{d}x}\right)\;\mathrm{d}x &amp;= 2x+1\end{align*}$$  Then we would have:     $$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\int f(x)\mathrm{d}x\right) = f(x) = \int\left(\dfrac{\mathrm{d}}{\mathrm{d}x}f(x)\right)\mathrm{d}x$$   So what is wrong with my thinking, and why isn't this the used definition?
20041
A function is "differentiable" if it has a derivative. A function is "continuous" if it has no sudden jumps in it.  Until today, I thought these were merely two equivalent definitions of the same concept. But I've read some stuff today which seems to be claiming that this is not the case.  The obvious next question is "why?" Apparently somebody has already asked:    Several answers were given, but I don't understand any of them. In particular, Wikipedia and one of the replies above both claim that $|x|$ has no derivative. Can anyone explain this extremely unexpected result?  Edit: Apparently some people dislike the fact that this is non-obvious to me. To be clear: I am not saying that the result is untrue. (I'm sure many great mathematicians have analysed the question very carefuly and are quite sure of the answer.) I am saying that it is extremely perplexing. (As a general rule, mathematics has a habit of doing that. Which is one of the reasons why we demand proof of everything.)  In particular, can anyone explain precisely why the derivative of $|x|$ at zero is not simply zero? After all, the function is neither increasing nor decreasing, which ought to mean the derivative is zero. Alternatively, the expression  $$\frac{|x + a| - |x - a|}{a}$$  becomes closer and closer to zero as $a$ becomes closer to zero when $x=0$. (In fact, it is exactly zero for all $a$!) Is that not how derivatives work?  Several answers have suggested that the derivative is not defined here "because there would be a jump in the derivative at that point". This seems to assert that a continuous function must never have a discontinuous derivative; I'm not convinced that this is the case. Can anyone confirm or refuse this argument?
61836
I've read many question/answer threads here on SE re: justification for the algebraic manipulation of $\mathrm{d}y/\mathrm{d}x$ in the standard formulation of calculus. I worked up my own shot at a justification using the definition of the derivative and would like to get the input of others whether it holds. It doesn't involve pushforward maps, non-standard analysis (two things I don't yet know), or any other tool than simple limit algebra/laws, the definition of the derivative, and (implicitly), the chain rule that makes the substitution possible. If it does hold as a justification as far as it goes, i.e., in one-dimension, it would seem to me, at least, to be a simple and clear justification in the form of an example that could be useful to other beginners. If goes wrong somewhere please let me know for my own understanding.   Integrate $\int \frac{x}{1+x^2}\mathrm{d}x$.  We'll do this by substitution. We want to restate the original problem in terms of $u$. That means that we want to state $x$, $1+x^2$, and $\mathrm{d}x$ in terms of $u$. Let $u=1+x^2 \rightarrow \frac{\mathrm{d}u}{\mathrm{d}x}=2x \rightarrow \mathrm{d}x=\frac{\mathrm{d}u}{2x}$. Let's justify that last step as that's where the justification (that I speak of) is required:  By definition, $\frac{\mathrm{d}u}{\mathrm{d}x}=2x$ is defined as $$ \lim_{\Delta x \to 0}\left( \frac{u(x+\Delta x)-u(x)}{\Delta x} \right)=2x $$ Recall that the limit of a quotient is equal to the quotient of the limits as long as the denominator is not equal to zero, thus the foregoing becomes $$ \frac{\lim_{\Delta x \to 0}\left( u(x+\Delta x)-u(x) \right)}{\lim_{\Delta x \to 0} \Delta x } =2x $$ Multiplying both sides by $\lim_{\Delta x \to 0}\Delta x$ and then dividing both sides by $2x$ yields $$ \frac{\lim_{\Delta x \to 0}\left( u(x+\Delta x)-u(x) \right)}{2x} =\lim_{\Delta x \to 0} \Delta x $$ which, in the limit, is equal to $$ \frac{\mathrm{d}u}{2x}=\mathrm{d}x $$ Rearranging we have $$ dx=\frac{\mathrm{d}u}{2x} $$ as desired.
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?
45377
I have been forewarned about it, I have read the  here, but I haven't seen a counter example where it doesn't work. I know that it isnt really a fraction, but does it effectively get the same result in all cases, or are there counterexamples I must be warned about.   I do not mean obvious counter examples, like $$\frac{dy}{dx}+\frac{du}{dv} = \frac{dydv+dxdu}{dx dv}$$ which as far as I know doesn't really mean anything.
8817
$dx$ appears in differential equations, such us derivatives and integrals.  For example, a function $f(x)$ its first derivative is $\dfrac{d}{dx}f(x)$ and its integral $\displaystyle\int f(x)dx$. But I don't really understand what $dx$ is.
40778
In integrals, we often make 'u' substitutions, which involve writing $u=f(x)$, and then also writing $du=f'(x)dx$. How valid is this? I've heard that you aren't allowed to do things like 'break' up derivatives like they do to solve first order differential equations. So can you do this with integrals, or is it also a mnemonic?
99618
While I do know that $\frac{dy}{dx}$ isn't a fraction and shouldn't be treated as such, in many situations, doing things like multiplying both sides by $dx$ and integrating, cancelling terms, doing things like $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ works out just fine.  So I wanted to know: Are there any particular cases (in single-variable calculus) we have to look out for, where treating $\frac{dy}{dx}$ as a fraction gives incorrect answers, in particular, at an introductory level?  Note: Please provide specific instances and examples where treating $\frac{dy}{dx}$ as a fraction fails
19533
The solution of differential equation $\displaystyle \left(\frac{dy}{dx}\right)^2-x\frac{dy}{dx}+y=0$   $\bf{My\; Try::}$ Let $\displaystyle \frac{dy}{dx} = t\;,$ Then Diferential equation convert into $t^2-xt+y=0$  So Its solution is given by $\displaystyle t=\frac{x\pm \sqrt{x^2-4y}}{2}$  So we get $$\frac{dy}{dx} = \frac{x\pm \sqrt{x^2-4y}}{2}$$  Now How can I solve after that, Help me  Thanks
27524
My mother is teaching a high school course on multivariable calculus, and they were studying linear differential equations of the form $$y' + P(x) y = Q(x),$$ and the question of why this equation is called "linear" came up.  In terms that these students are familiar with, since they haven't been exposed to linear algebra yet, my thought was to say that the equation for a line, $y = mx+b$, is "linear" in $x$ (ignoring the technicality that it's actually an affine equation, not a linear one), because it's in the form "coefficient times $x$", and then we allow another term which is just a lonely coefficient. And then we extend this notion to saying that the above differential equation is "linear" in $y$ and $y'$, but this time the coefficients are allowed to be functions of $x$.  That's probably a good enough hand-wavy explanation to help students remember the definition, at the very least. I couldn't really think of a good reason why it "should", a priori, be kosher to allow coefficients to be functions of $x$ here. At that point it seems to me like you just have to get into the linear algebraic definition of linearity, which, being completely foreign to the students... it just seems to be a bit too deep of a rabbit hole for this purpose.  So my question is: does anyone have a better way of approaching this? And if you think my hand-wavy explanation above is largely acceptable, is there a way you can explain why multiplying by non-constant functions of $x$ "should" still be considered linear in $y$?