5
$\begingroup$

I'm not sure if I derived it right. So, Let $f(x)$ = $(\sin x)^x$. $f' = x(\cos x)$ right? then just substitute so $f' = \frac{\pi}{2}(\cos\frac{\pi}{2})$ ? Since $\cos(180°)$ is equal to $-1$. Then its -$\frac{\pi}{2}$? they said it's wrong.

after reading all the comments I arrived with $f'$ = $(\sin x)^x$ $\frac{\sin x\ln(\sin x) + x\cos x}{\sin x}$ then substitute $\frac{\pi}{2}$ to x so that's $90 \deg$ $\cos90$ is $0$ and $\sin90\deg$ is $1$ so $1$ raised to $90\deg$ then $\frac {(1)(1)+0}{1}$ my final answer is now $\frac{\pi}{2}$ which is in the multiple choice. But is it correct?

$\endgroup$
2
  • $\begingroup$ Is this function differentiable? Isn't the function discontinuous at 0? $\endgroup$ Oct 12, 2014 at 9:17
  • $\begingroup$ @kart I'm not I just answered this in a quiz and I got zero. I just applied the log. differentiation then substitute $90deg.$ to all x $\endgroup$
    – Mickey
    Oct 12, 2014 at 9:19

6 Answers 6

4
$\begingroup$

Given that the problem asks for the value of $f'$ at a specific point, rather than for a general formula for $f'$, we might look for a simple solution involving geometry, rather than differentiation rules.

Note that $0\leq\sin(x)\leq1$ on $[0,\pi]$, thus $f(x)=(\sin(x))^x$ is also between $0$ and $1$ on that interval. Furthermore, $f(\pi/2) = 1$. Thus, $f$ has a maximum at $\pi/2$ so that $f'(\pi/2)=0$.


Admittedly, I came up with this explanation after noticing that the answer was zero. Such a simple answer, though, only begs the question further - namely, is there a simple geometric explanation.

$\endgroup$
0
3
$\begingroup$

No, $f'(x)$ is not $x\cos x$. To find $f'$, first take natural logarithm of both sides of the definition of $f$: $$ \ln(f(x)) = \ln\left((\sin x)^x\right)=x\ln(\sin x) $$ Then take the derivative of both sides of the above equation. The left side becomes $\frac{f'(x)}{f(x)}=\frac{f'(x)}{(\sin x)^x}$, so this allows you to solve for what $f'(x)$ is.

$\endgroup$
7
  • $\begingroup$ What if $\sin x < 0$ ? $\endgroup$
    – Zhen Zhang
    Oct 12, 2014 at 8:58
  • $\begingroup$ so the answer is $f'$ = $\frac{xcosxsinx^x}{sinx}$ ? $\endgroup$
    – Mickey
    Oct 12, 2014 at 9:02
  • $\begingroup$ @Mickey Not quite. I think you made a mistake computing the derivative of the right hand side, $x\ln\sin x$. Be careful in your use of product rule/chain rule. $\endgroup$ Oct 12, 2014 at 9:07
  • $\begingroup$ @ZhenZhang $f(x)$ is not a real-valued function when $\sin x<0$; for example, $f(-1/2)=1/\sqrt{\sin(-1/2)}$ is not real. In these cases, the method still works, but you have to use a complex branch of $\log$. $\endgroup$ Oct 12, 2014 at 9:10
  • $\begingroup$ @mike hmm um i think this is it $f'$ = $(sinx)^x$ $\frac{sinxln(sinx)+xcosx}{sinx}$ product rule my bad $\endgroup$
    – Mickey
    Oct 12, 2014 at 9:11
3
$\begingroup$
$\endgroup$
6
  • 2
    $\begingroup$ Try to avoid link only answers :) $\endgroup$ Oct 12, 2014 at 8:55
  • $\begingroup$ @TheGame That is a hint for the answer so that the OP can work on it. $\endgroup$
    – C.S.
    Oct 12, 2014 at 8:56
  • $\begingroup$ @s.c. so the answer is $f'$ = $\frac{xcosxsinx^x}{sinx}$ ? $\endgroup$
    – Mickey
    Oct 12, 2014 at 9:04
  • $\begingroup$ @TheGame: it is not simply a link, it is suggested to take the log and even if you don't follow the link, differentiation is indicated. $\endgroup$
    – robjohn
    Oct 12, 2014 at 9:33
  • $\begingroup$ @S.C.: when giving a hint, it is a good idea to explicitly state that so that others know it is a hint. Hints seem to be judged differently than full answers. $\endgroup$
    – robjohn
    Oct 12, 2014 at 9:34
2
$\begingroup$

$ f'(x) = xsin(x)^{x-1}*cos(x) + (sin(x))^{x}*ln(sin(x))$ Now $f'(pi/2)$ is the limit as x goes to 0 of $f'(x)$. The first term is 0 and $sin(x)^{x}$ is 1 when x approaches 0. Therefore $f'(pi/2)$ is lim as x -> 0 of $ln(sin(x))$ which is $-\infty$.

$\endgroup$
1
  • 1
    $\begingroup$ this might help you :) $\endgroup$
    – Bman72
    Oct 12, 2014 at 9:24
2
$\begingroup$

Starting with $f(x)=\sin(x)^x$ note that $\ln|f(x)|=x\ln|\sin(x)|$ (the absolute values solve negativity problems) and the derivative of this is $$\frac{f'(x)}{f(x)}=\ln|\sin(x)|+\frac {x\cos(x)}{\sin(x)}$$

Plugging in $x=\frac\pi2$, we get $$f'\left(\frac\pi2\right)=0$$

$\endgroup$
0
$\begingroup$

use that $f(x)=(\sin(x))^x=e^{x\ln(\sin(x))}$ thus we get $f'(x)=e^{x\ln(\sin(x))}(\ln(\sin(x))+x\frac{\cos(x)}{\sin(x)})$

$\endgroup$
8
  • 2
    $\begingroup$ $a^b=e^{b\ln(a)}$ $\endgroup$ Oct 12, 2014 at 8:54
  • 2
    $\begingroup$ Oh, that's not right $\endgroup$
    – robjohn
    Oct 12, 2014 at 9:12
  • $\begingroup$ what do you mean? $\endgroup$ Oct 12, 2014 at 9:13
  • $\begingroup$ it is right, take the logarithm of both sides $\endgroup$ Oct 12, 2014 at 9:14
  • 1
    $\begingroup$ $(\sin(x))^x=e^{x\log(\sin(x))}$ $\endgroup$
    – robjohn
    Oct 12, 2014 at 9:20

You must log in to answer this question.

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