0
$\begingroup$

$(1)\quad $ How to differentiate $y=\frac 12 \sin x$?

I know that $\frac{dy}{dx}$ of $y=\sin x$ is easy to calculate: $\frac{dy}{dx} = \cos x$. But what if there is a coefficient preceding before it?

$(2)$ Another question is: Does the following equality of functions hold? $$\frac{x(x^2+1)}{x^2+1}\overset{?}{=}x$$

Please explain this because I know that function $\dfrac{x\cdot x}{x}$ does not equal function $x$ (the first function is not defined at $0$ but the second one is defined at $0$).

But the question I asked, the domain of the function can be any number according to what I think because the denominator cannot be zero, it's always a positive number $x^2+1$.

$\endgroup$
  • $\begingroup$ To your second question, since $x^2+1$ is never $0$ (at least in the real numbers) $\frac{x^2+1}{x^2+1}=1$ and $\frac{x(x^2+1)}{x^2+1} = x$ for all $x$. So yes, the functions are equal. $\endgroup$ – Thomas Andrews Sep 11 '14 at 13:10
  • $\begingroup$ Do you mean $\frac{1}{2\sin x}$ or $\frac{1}{2}\sin x$? $\endgroup$ – Thomas Andrews Sep 11 '14 at 13:11
  • $\begingroup$ I mean 1/2 (sin x) $\endgroup$ – off99555 Sep 11 '14 at 13:13
  • 1
    $\begingroup$ One should avoid different questions in a same question on MSE. $\endgroup$ – user37238 Sep 11 '14 at 13:16
  • $\begingroup$ This is my first time using this website, I will be aware next time. $\endgroup$ – off99555 Sep 11 '14 at 13:18
3
$\begingroup$

Question $(1):\quad$ Given $y = \dfrac 12 \sin x$:

$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac 12 \sin x\right) = \frac 12\cdot \frac{d}{dx}(\sin x) = \frac 12 \cos x$$

$$y = a f(x)\implies \frac{dy}{dx} = \frac{d}{dx}(a f(x)) = af'(x)$$ for all constants $a$.


Question $(2):$

$$\frac {x(x^2 + 1)}{x^2 + 1 } = x \quad \forall x \in \mathbb R$$

This happens to be the case in this example because there are no real values of $x$ at which the left-hand side is undefined. Specifically, as you note, $x^2 + 1>0$ for all $x$, and hence the function is defined everywhere. So we may cancel the common factor $x^2 + 1$ without changing the function in any way.

In the case of $f(x) = \frac{x^2}{x}$, which you refer to, here we do have problems with simply canceling a common factor of $\,x.\,$ Specifically, $\,\dfrac{x^2}{x}\,$ is undefined at $x = 0$, whereas the function $g(x) = x$ is defined everywhere, so the functions are not equivalently defined, i.e., they are not equivalent functions.

$\endgroup$
  • $\begingroup$ I would say that the first response is missing one sign "=" -> ambiguity. $\endgroup$ – georg Sep 11 '14 at 14:08
  • $\begingroup$ Thanks, @georg. I didn't catch that until you mentioned it.. $\endgroup$ – amWhy Sep 11 '14 at 14:19
2
$\begingroup$

For the first part of your question, you can remove constants when differentiating:

$$\frac{\mathrm{d}}{\mathrm{d}x}\big(a\cdot f(x)\big)=a\frac{\mathrm{d}}{\mathrm{d}x}f(x)$$

You can prove this with the product rule by finding $a\frac{\mathrm{d}}{\mathrm{d}x}f(x)+f(x)\frac{\mathrm{d}}{\mathrm{d}x}a$ & noting that the derivative of a constant is $0$. It implies that your derivative is $\frac{\mathrm{d}}{\mathrm{d}x}\big(\frac{1}{2}\sin(x)\big)=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)$.

For the second part of the question, it is valid to cancel terms in a function but just be aware that the function's domain should stay the same. If the original function is undefined at a number, it should stay that way. As Thomas Andrews points out, $x^2+1$ is never $0$ for real $x$ so you can effectively cancel the $\frac{(x^2+1)}{(x^2+1)}$ term.

$\endgroup$
  • $\begingroup$ I got the edit wrong - he wanted $y=\frac{1}{2}\sin x$. Sorry. $\endgroup$ – Thomas Andrews Sep 11 '14 at 13:14
  • $\begingroup$ @ThomasAndrews Ah, thanks for letting me know :) No problem by the way, I've done it before too. $\endgroup$ – Jam Sep 11 '14 at 13:14
  • $\begingroup$ Thank you very much for your help. I would like to vote both of you up but I can't. My reputation is not high enough to do that. Both answers are very useful for me. But I don't understand what setting a as a function of x mean. Can you show me please? $\endgroup$ – off99555 Sep 11 '14 at 14:25
  • $\begingroup$ @off99555 Glad I could help. What I meant by "setting it as a function" was that, by using the product rule, we can find the derivative of $f(x)\cdot g(x)$ as $f(x)g'(x)+g(x)f'(x)$. If we say that $g(x)$ is a constant $a$, we could find the derivative of $a\cdot f(x)$. Since the derivative of any constant is $0$, it shows us that we can take out the constant. So $\frac{d}{dx}(\frac{1}{2}\sin(x))=\frac{1}{2}\frac{d}{dx}\sin(x)$. Does that make sense? $\endgroup$ – Jam Sep 11 '14 at 14:45
  • $\begingroup$ Sure. That make sense! That's what I'm looking for, the fundamental explanation. $\endgroup$ – off99555 Sep 17 '14 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.