2
$\begingroup$

I am suppose to differentiate

$y=(\sin x)^{\ln x}$

I have absolutely no idea, this was asked on a test and I just do not know how to do this I have forgotten the tricks I was suppose to memorize for the test.

  • 3
    If you take the logarithms of both sides, you have $\ln\,y=(\ln\,x)(\ln\sin\,x)$. You can use implicit differentiation, the chain rule, and the product rule from here.2012-05-04
  • 0
    I don't have those things memorized anymore. Also I don't know where the lnx went.2012-05-04
  • 2
    You may not have them memorized, but you can look them up: [implicit differentiation](http://en.wikipedia.org/wiki/Implicit_differentiation#Implicit_differentiation), the [chain rule](http://en.wikipedia.org/wiki/Chain_rule) and the [product rule](http://en.wikipedia.org/wiki/Product_rule).2012-05-04
  • 0
    I am trying to review for my final without looking things up.2012-05-04
  • 0
    J.M. rewrote the equation with the power property of ln. Implicit differentiation would have you take d/dy of both sides, and taking the derivative of the right side would require the power rule and chain rule. Come back with your progress and we can help you where you get stuck.2012-05-04
  • 2
    Unsolicited advice: "memorizing" everything, and "reviewing without looking things up" are generally bad study policies.2012-05-04
  • 0
    Well so is looking things up, it only leads me to forget everything on the test.2012-05-04
  • 0
    I think I get $\frac{y\prime}{y} = \frac{lnsinx}{x} + \frac{lnx cosx}{sinx}$2012-05-04
  • 0
    Looks good: solve for $y'$ and eliminate the $y$! Consider writing up the solution in an Answer box below. There's no rule against answering your own questions, especially if you're looking for feedback about the solution.2012-05-04
  • 0
    Would it be wrong to rewrite y as $sinx^{lnx}$?2012-05-04
  • 0
    @Jordan No, $(\sin x)^{\ln x}=\sin^{\ln x}x\ne \sin x^{\ln x}$2012-05-04
  • 0
    @AméricoTavares I do not understand the difference.2012-05-04
  • 1
    @Jordan $y=(\sin x)^{\ln x}$ means that $y$ is $\sin x$ raised to the $\ln x$ power, while $\sin x^{\ln x}=\sin (x^{\ln x})$.2012-05-04

2 Answers 2

5

Hint

$$\sin x=e^{\ln \left( \sin x\right) }\Rightarrow \left( \sin x\right) ^{\ln x}=\left( e^{\ln \left( \sin x\right) }\right) ^{\ln x}=e^{\left( \ln x\right) \;\cdot\;\ln \left( \sin x\right) }\tag{1}$$

and evaluate the derivative of $e^{\left( \ln x\right) \ln \left( \sin x\right) }.$

Comments (trying to reply to OP's comments).

  1. We can start by writing the given function as $$y=(\sin x)^{\ln x}=e^{\left( \ln x\right) \;\cdot\;\ln \left( \sin x\right)},\tag{2}$$ which is a particular case of the algebraic identity $$\left[ u\left( x\right) \right] ^{v\left( x\right) }=e^{v(x)\;\cdot\;\ln(u(x))}.\tag{3}$$ Remarks. We've used the following properties. By the definition of the natural logarithm, we have (see Powers via logarithms) $$\ln u=v\Leftrightarrow u=e^v=e^{\ln u},\tag{4}$$ and the rule $(a^b)^c=a^{b\;\cdot\; c}\tag{5}$
  2. Finally we evaluate the derivative of $e^{g(x)}$, where $$g(x)=\left( \ln x\right)\;\cdot\; \ln \left( \sin x\right)\tag{6}.$$ By the chain rule we have $$y'=(e^{g(x)})'=e^{g(x)}g^{\prime }(x),\tag{7}$$ and $g'(x)$ is to be computed by the product rule.

Evaluation of $(7)$ $$\begin{eqnarray*} g^{\prime }(x) &=&\left( \left( \ln x\right) \ln \left( \sin x\right) \right) ^{\prime } \\ &=&\left( \ln x\right) ^{\prime }\ln \left( \sin x\right) +\left( \ln x\right) \left( \ln \left( \sin x\right) \right) ^{\prime } \\ &=&\frac{1}{x}\ln \left( \sin x\right) +\left( \ln x\right) \frac{\cos x}{ \sin x} \\ &=&\frac{\ln \left( \sin x\right) \sin x+\left( \left( \ln x\right) \cos x\right) x}{x\sin x}.\tag{8} \end{eqnarray*}$$

Hence, since $e^{g(x)}=\left( \sin x\right) ^{\ln x}$, we obtain $$\begin{eqnarray*} y^{\prime } &=&e^{g(x)}g^{\prime }(x)=y\;\cdot\; g^{\prime }(x) \\ &=&\left( \sin x\right) ^{\ln x}\frac{\ln \left( \sin x\right) \sin x+\left( \left( \ln x\right) \cos x\right) x}{x\sin x}.\tag{9} \end{eqnarray*}$$

  • 0
    I do not follow what you did.2012-05-04
  • 1
    $e^x$ and $\ln x$ are inverse functions. Therefore, $e^{\ln u}=u$. Then for the second part, $(a^b)^c=a^{bc}$. The function should now be in a form where you can take the derivative.2012-05-04
  • 0
    @Jordan: I used the definition of the natural logarithm, and the algebraic rule for a power of a power.2012-05-04
  • 0
    @Mike Thanks for the comment.2012-05-04
  • 0
    So $e^{lnx}$ = $e^x$ = $lnx$?2012-05-04
  • 1
    @Jordan No, $e^{\ln x}=x$2012-05-04
  • 0
    I am having trouble following what is happening in your answer, you have a transformation of sinx into other things but I do not understand where any of it comes from.2012-05-04
  • 1
    @Jordan I've thought evaluating the derivative of $e^{\text{some function of } x}$. So I converted the given function $y=(\sin x)^{\ln x}$ to that form. I started by writing $\sin x$ as $e^{\ln(\sin x)}$.2012-05-04
  • 0
    @AméricoTavares I am a little confused where the lnx goes.2012-05-04
  • 1
    @Jordan $\ln x$ is the exponent of $\sin x$. So, after converting $\sin x$ to $e^{\ln (\sin x)}$, we have to multiply the exponents $\ln x$ and $\ln (\sin x)$.2012-05-04
  • 0
    Why is it level to raise everything to a power of e? Are they always equal statements? So I have lnsinx because of some power rule with logs?2012-05-04
  • 1
    @Jordan Because of the power rule $(u^v)^w=u^{v\cdot w}$ mentioned in 2, applied to $u=e, v=\ln(\sin x), w=\ln x.$2012-05-04
  • 1
    @Jordan An inverse function $f^{-1}(u)$ is a function such that $f^{-1}(f(u))=f(f^{-1}(u))=u$. Since $\ln u$ and $e^u$ are inverse functions, $e^{\ln u}=\ln e^u=u$. Now replace u with $\sin x$. Then multiply exponents.2012-05-04
  • 0
    @Jordan Do you want me to rewrite this answer, instead of leaving it as it is with all the edits?2012-05-04
  • 0
    @AméricoTavares I think it is good I will just have to go through it a couple times to comprehend it fully.2012-05-04
  • 0
    @Jordan In the meantime I rewrote it. You you need I'll roll it back.2012-05-04
3

You can either use the definition that says that $$a^b = e^{b\ln(a)}$$ and use the chain rule with $$y(x) = (\sin x)^{\ln x} = e^{(\ln(x))(\ln(\sin x))}$$ or you can use logarithmic differentiation.

If $y = (\sin x)^{\ln x}$, then taking logarithms on both sides we get $$\ln y = \ln\left((\sin x)^{\ln x}\right) = (\ln x)\ln(\sin x).$$ Now using implicit differentiation we have: $$\begin{align*} \frac{d}{dx}\ln y &= \frac{d}{dx}\left( (\ln x)\ln(\sin x)\right)\\ \frac{1}{y}\frac{dy}{dx} &= \left(\ln x\right)'\ln(\sin x) + (\ln x)\left(\ln (\sin x)\right)'\\ \frac{y'}{y} &= \frac{1}{x}\ln(\sin x) + (\ln x)\left(\frac{1}{\sin x}(\sin x)'\right)\\ \frac{y'}{y} &= \frac{\ln\sin x}{x} + \frac{(\ln x)\cos x}{\sin x}\\ \frac{y'}{y} &= \frac{\ln \sin x}{x} + \ln x\cot x\\ y' &= y\left(\frac{\ln \sin x}{x} + \ln x\cot x\right)\\ y' &= (\sin x)^{\ln x}\left(\frac{\ln \sin x}{x} + \ln x\cot x\right). \end{align*}$$