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I have to solve a third order differential equation.

Not the topic of question, but I was thinking of letting $f(x)$ (which I'm solving for) be equal to $y$, so that I can use $dy$ and $dx$.

If I didn't involve y, would I be using $df$ and $dx$, or $df(x)$ and $dx$, or what, would be my differentials?

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  • $\begingroup$ $\frac {dy}{dx}, \frac {df}{dx}, \frac {du}{dx}, \frac {dx}{dt}\cdots$ you will see all of them. It doesn't really matter what the letters happen to be. $\endgroup$ – Doug M Mar 28 '17 at 5:43
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You wouldn't be confused so much with representation if you think in terms of what you are actually doing while we find differentials.

For any equation (basically a curve in your graph), you are finding the slope(rise/fall) to figure out where it attains min and max and the curves behavior.

When you do, $\frac{d(x^2)}{dx}$ or $\frac{dy}{dx}$ where $f(x)=x^2,$ or $y=x^2.$ It's all the same. You are just figuring out how the slope of the curve changes with respect to changing values in $x$ axis.

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  • $\begingroup$ i'm not confused about what to do, I just don't know the proper notation for differentials $\endgroup$ – Saketh Malyala Mar 29 '17 at 2:01
  • $\begingroup$ df/dx is not right in my opinion. It doesn't tell much. Because only f(x) is a function in x. $\endgroup$ – Komal-SkyNET Mar 29 '17 at 4:35
  • $\begingroup$ That's why I had mentioned in my answer that you wouldn't be confused if you knew what is going on when you are differentiating an equation, you need something in denominator that shows against which axis your changing values are- in this case dx. In numerator you need an equation that's in x. Like d(x^2)/dx. And now you might see why df/dx is wrong. f is not equal to x^2 whereas f(x)=x^2. $\endgroup$ – Komal-SkyNET Mar 29 '17 at 4:41
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It must be $$\frac{d(f(x))}{dx}$$ Since, $$\frac{dy}{dx}=\frac{d(f(x))}{dx}$$

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  • $\begingroup$ Would df be acceptable as well? $\endgroup$ – Saketh Malyala Mar 28 '17 at 5:45
  • $\begingroup$ I don't think so. Since $f$ is a function not a variable. $\endgroup$ – Harsh Kumar Mar 28 '17 at 14:36
  • $\begingroup$ Since in my answer, I just substitute $y=f(x)$ $\endgroup$ – Harsh Kumar Mar 28 '17 at 14:37

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