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Finding the differential of $x \sin(x)$ \begin{align} d(x \sin x) &= (dx) sinx + x (d(\sin x)) &&\text{using the product rule}\\ &= \sin x\,dx + x \cos x\,dx \end{align}

My question is why is the differential of $\sin x$ equal to $(\cos x) (dx)$ and not just $\cos x$ (which is the derivative of $\sin x$.)

A similar question is, why is the differential of $x^2$ equal to $2x(dx)$? Why is there an additional $dx$ in both cases?

egreg
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Minu
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    For purposes of a calculus class, you should just take this as a definition: $dy = \frac{dy}{dx} dx$. The notation is motivated by the fact that if $\Delta x$ is small, $\Delta y = \frac{\Delta y}{\Delta x} \Delta x \approx \frac{dy}{dx} \Delta x$. – Ian Jul 11 '14 at 21:33
  • just divide by $dx$ – Alex Jul 11 '14 at 21:37
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    If you believe in using differentials as algebraic symbols, then yeah, you can go from $\frac{dy}{dx} = \frac{dy}{dx}$ to $dy = \frac{dy}{dx} dx$ by multiplying both sides by $dx$. I think it is probably better for calculus students to not get confused by this, and just take $dy = \frac{dy}{dx} dx$ as a formal, symbolic definition. – Ian Jul 11 '14 at 21:49
  • Do you mean that because d/dx (sin x) = cos x, d(sin x) = cos x(dx)? – Minu Jul 11 '14 at 21:52

4 Answers4

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in short: $f'(x) = \frac{d f(x)}{dx}$. In your case $f(x) = x \sin x$.

Alex
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You have a function $f(x)$ and are looking for the differential $df(x)$. In the case $f(x)=x\sin(x)$:

$$\begin{align} f'(x) &= \frac{df(x)}{dx}\\ df(x) &= f'(x)dx\\ df(x) &= \left( \frac{df(x)}{dx} \right) dx \\ d\left( x \sin (x) \right) &=\left( x \cos(x) + \sin(x) \right)dx \end{align} $$

Similarly, for $f(x)=x^2$: $$\begin{align} f'(x) &= \frac{df(x)}{dx}\\ df(x) &= f'(x)dx\\ df(x) &= \left( \frac{df(x)}{dx} \right) dx \\ d(x^2) &= (2x)dx \end{align} $$

druckermanly
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Here we have to do with infinitesimals.

What is the derivative? We all know the formula: $$\frac{\Delta f(x)}{\Delta x}=f'(x).$$ This means that derivative is a value that tells us the ratio between the change of the function and the change of its variable. If you have a knowledge in numerical analysis, roughly speaking this is similar to relative error.

But what if we want to know the value of change for the function f(x)? Then, from the last formula we may write: $$\Delta f(x)=f'(x) \Delta x.$$ So this tells us that the change of $f(x)$ is equal to the product of $f'(x)$ and the change of the variable corresponding to the change of the function. In 'numerical analysis language' this is a kind of absolute error of the function $f(x)$.

In the end, we write $d$ instead of $\Delta$ when we deal with the limit.

Emo
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It's precisely because you're asking for the differential, rather than asking for the derivative with respect to $x$. In general,

$$ \mathrm{d}(f(x)) = f'(x) \mathrm{d} x$$

One interpretation of taking "the derivative with respect to $x$" is to first compute the differential, and then substitute $\mathrm{d} x = 1$. I imagine your confusion is because you are so used to that final substitution that you don't even realize it's a thing. But here, we're specifically asking for the differential, so you don't make that substitution.

(this interpretation is more like taking a partial derivative than taking a total derivative, if you know what those phrases mean)

This is something that is useful, because we may not yet have decided that $\mathrm{d} x = 1$ is what we want. Maybe we will want to substitute $x = g(t)$ and eventually substitute $\mathrm{d} t = 1$.

Or maybe you simply want to explicitly record how the function varies.

Notation with differential is especially useful when you have more than one variable (even if you only have one independent variable).

  • We do not substitute dx=1 when we deal with the derivative. In that case we have a 'fraction' of infinitesimals df(x) and dx. – Emo Jul 11 '14 at 22:44
  • @Emin: It depends on whether you're thinking of a derivative as being like a partial derivative but just with one variable. (I usually don't -- I usually think of it as a ratio of differentials) And the interpretation of substituting $\mathrm{d}x$ with something is directly relevant to various applications, such as differential approximation. –  Jul 11 '14 at 22:46
  • Partial derivative? I don't understand what do you mean with partial derivative! We are dealing with functions with one variable. – Emo Jul 11 '14 at 22:53