1
$\begingroup$

What is the meaning of $d(sin(x))$?

like say you have an integral containing $cos(x)$ as a function and you want to substitute:

$d(sin(x))=cos(x)dx$

I don't understand what $d(sin(x))$ is and how you would manipulate it? Thanks.

$\endgroup$
4
  • $\begingroup$ Have you seen trigonometric substitutions used to solve integration problems? Would an example of this help to understand how it can be useful? $\endgroup$
    – hardmath
    Commented Oct 13, 2016 at 19:19
  • $\begingroup$ I have just been on google trying to understand this, does this simply mean "the derivative of the sine of x = the cosine of x w.r.t x"? $\endgroup$
    – Andy6978
    Commented Oct 13, 2016 at 19:22
  • 1
    $\begingroup$ That is closely related but you are asking about a differential rather than a derivative. $\endgroup$
    – hardmath
    Commented Oct 13, 2016 at 19:25
  • $\begingroup$ ah okay, thanks. $\endgroup$
    – Andy6978
    Commented Oct 13, 2016 at 19:38

1 Answer 1

1
$\begingroup$

It's the exterior derivative of the function $\sin x$.

If $f$ is a function of $x$, then how does a change in $x$ cause $f(x)$ to change?

In general: $\Delta f \approx f'(x)\cdot \Delta x$, where $\Delta f$ is the change in $f(x)$ and $\Delta x$ is the change in $x$.

In your case $\Delta(\sin x) \approx \cos x \cdot \Delta x$.

As $\Delta x \to 0$, we write $\mathrm df = f'(x)~\mathrm dx$ and $\mathrm d(\sin x) = \cos x~\mathrm dx$.

This is an example of a differential one form, and these can be integrated.

If $f$ is a function of two variables, say $x$ and $y$, then

$$\Delta f \approx \frac{\partial f}{\partial x}~\Delta x+\frac{\partial f}{\partial y}~\Delta y$$ gives an approximation of how $f$ changes, given small changes to $x$ and $y$.

As $\Delta x \to 0$ and $\Delta y \to 0$ then we write $$\mathrm df= \frac{\partial f}{\partial x}~\mathrm d x+\frac{\partial f}{\partial y}~\mathrm d y$$

$\endgroup$
1
  • 1
    $\begingroup$ There is nothing in the OP about multivariable functions, nor about "higher degree" forms (differentiation). From your first link: "On a differentiable manifold, the exterior derivative extends the concept of the differential of a function to differential forms of higher degree." Thus the introduction of this exterior derivative concept seems like overkill. $\endgroup$
    – hardmath
    Commented Oct 13, 2016 at 23:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .