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Okay this may sound stupid but I need a little help... What do $\Large \frac{d}{dx}$ and $\Large \frac{dy}{dx}$ mean?

I need a thorough explanation. Thanks.

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    $\begingroup$ This is not precalculus, it is calculus. These symbols are derivatives. Are you familiar with derivatives?\ $\endgroup$ Commented Mar 25, 2013 at 15:44

5 Answers 5

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The symbol $$ \frac{dy}{dx} $$ means the derivative of $y$ with respect to $x$. If $y = f(x)$ is a function of $x$, then the symbol is defined as $$ \frac{dy}{dx} = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}. $$ and this is is (again) called the derivative of $y$ or the derivative of $f$. Note that it again is a function of $x$ in this case. Note that we do not here define this as $dy$ divided by $dx$. On their own $dy$ and $dx$ don't have any meaning (here). We take $\frac{dy}{dx}$ as a symbol on its own that can't be slit up into parts.

The symbol $$ \frac{d}{dx} $$ you can consider as an operator. You can apply this operator to a (differentiable) function. And you get a new function. So if $f$ is a (differentiable) function that it makes sense to "apply" $\frac{d}{dx}$ to $f$ and write $$ \frac{d}{dx}f $$ If you write $y = f(x)$, then this is the same as $$ \frac{d}{dx}y = \frac{dy}{dx}. $$

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    $\begingroup$ The confusion often arises from the fact that many writers call $dy/dx$ a "symbol" as if it were atomic, but then later start doing algebra with it. This leads to the question, "well, then what is $dy$ really?" $\endgroup$
    – Fixee
    Commented Mar 25, 2013 at 15:55
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    $\begingroup$ @Fixee: That is right. I often see this confusion which is why I always emphasize that $\frac{dy}{dx}$ is just a symbol. It is not a fraction. $\endgroup$
    – Thomas
    Commented Mar 25, 2013 at 15:57
  • $\begingroup$ @Thomas It's kinda a fraction. It's the limit of a fraction. $\endgroup$ Commented May 1, 2015 at 21:35
  • $\begingroup$ (Or the "standard part" of a fraction, if you're doing nonstandard analysis...) $\endgroup$ Commented May 1, 2015 at 21:35
  • $\begingroup$ @Fixee See my answer. It is nevertheless CAN be defined as a fraction of two functions, rather than an atomic object. $\endgroup$
    – porton
    Commented May 1, 2015 at 21:59
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$\frac{d}{dx}$ means differentiate with respect to $x$.

$\frac{dy}{dx}$ means differentiate $y$ with respect to $x$.

Do you have any concrete examples for which you need to calculate these two? It would probably make it more easy to grasp for you if I could explain it in a few examples.

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  • $\begingroup$ Maybe calculating the flow rate of a tap based on how the container gets filled? $\endgroup$ Commented Mar 23, 2023 at 7:57
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$d f$ means the differential of function $f$. By definition $(df)(x) = \lambda t\in\mathbb{R}:f'(x)\cdot t$. In other words, differential is the linear function (of an additional variable denoted $t$ here) whose tangent is the derivative of $f$.

$d$ alone means the differential operator (a function of argument $f$).

Exercise: Show that $\frac{df}{dx}=f'$.

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  • $\begingroup$ Note that $\frac{F}{G}$ is defined for two functions $F$ and $G$ as $\frac{F}{G}(x)=\frac{F(x)}{G(x)}$. Thus $\frac{df}{dx}$ makes sense. $\endgroup$
    – porton
    Commented May 1, 2015 at 21:16
  • $\begingroup$ It is described in en.wikipedia.org/wiki/Differential_of_a_function $\endgroup$
    – porton
    Commented May 2, 2015 at 19:33
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I like to look at it this way: $dx$ and $dy$ are just representations of change in accordance to either $x$ or $y$ axis. If you take the the symbol for derivative $$\frac{dy}{dx}$$ and compare it to the formula for the slope: $$\frac{f(x_1) - f(x_2)}{x_1 - x_2}$$ we can clearly see that $dy$ and $dx$ depict change in $y$ and change in $x$ respectively.

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If $y=f(x)$ i.e., where $y$ is the equation ( the dependent variable) and $x$ is the independent variable. Meaning $x$ changes $y$.

Now $\frac{dy}{dx}$ means differentiate the equation $y$ in respect to $x$.

$\frac{d}{dx}$ means differentiate in respect to $x$.

Same way $\log x$ means find the natural logarithm of $x$, $\frac{d}{dx} x$ means find the derivative of $x$.

N.B. In an equation $k= h²+5 $, $\frac{dk}{dh}$ means differentiate the equation $k$ in respect to $h$. Its not always $\frac{dy}{dx}$.

I hope you understand

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  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, use MathJax. $\endgroup$
    – dantopa
    Commented Jun 27, 2019 at 3:57
  • $\begingroup$ See the answer given by Thomas $\endgroup$
    – nmasanta
    Commented Jun 27, 2019 at 4:52

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