 # Trouble with Logarithmic Differentiation Hey guys I'm trying to find the derivative of this equation using logarithmic differentiation but I'm having some trouble. Wolfram Alpha is giving me different answers and I'm having difficulty understanding the steps it spits out. Could somebody take a look at what I'm doing and tell me where I went wrong? The problem I need to differentiate is: y=(sinx)^(ln x) My first step is to take the logarithm of both sides and factor out the exponent. ln(y)=lnx∗ln(sinx) Next I try to differentiate both sides, using the rules for logarithms: (1)/(y)*y'=(1)/(x)(1)/(sin x)cos x =>(1)/(y)*y'=(cos x)/(x sin x) Which finally renders: (dy)/(dx)=(cos x)/(x sin x)(sin x)^(ln x) But this doesn't seem to be the right answer, could somebody help me out? Stephanie Hunter 2022-07-20 Answered
Trouble with Logarithmic Differentiation
Hey guys I'm trying to find the derivative of this equation using logarithmic differentiation but I'm having some trouble. Wolfram Alpha is giving me different answers and I'm having difficulty understanding the steps it spits out. Could somebody take a look at what I'm doing and tell me where I went wrong?
The problem I need to differentiate is:
$y=\left(\mathrm{sin}x{\right)}^{\mathrm{ln}x}$
My first step is to take the logarithm of both sides and factor out the exponent.
$\mathrm{ln}\left(y\right)=\mathrm{ln}x\ast \mathrm{ln}\left(\mathrm{sin}x\right)$
Next I try to differentiate both sides, using the rules for logarithms:
$\frac{1}{y}\ast {y}^{\prime }=\frac{1}{x}\frac{1}{\mathrm{sin}x}\mathrm{cos}x⇒\frac{1}{y}\ast {y}^{\prime }=\frac{\mathrm{cos}x}{x\mathrm{sin}x}$
Which finally renders:
$\frac{dy}{dx}=\frac{\mathrm{cos}x}{x\mathrm{sin}x}\left(\mathrm{sin}x{\right)}^{\mathrm{ln}x}$
But this doesn't seem to be the right answer, could somebody help me out?
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it constanzma
You need to use the product rule when differentiating:
$\frac{d}{dx}\left(fg\right)=\frac{df}{dx}g+f\frac{dg}{dx}$
So if $\mathrm{ln}y=\mathrm{ln}x\cdot \mathrm{ln}\left(\mathrm{sin}x\right)$, for the purposes of differentiating the right hand side we have that $f=\mathrm{ln}x$ and that $g=\mathrm{ln}\left(\mathrm{sin}x\right)$. So
$\frac{{y}^{\prime }}{y}=\frac{d}{dx}\left(\mathrm{ln}x\right)\cdot \mathrm{ln}\left(\mathrm{sin}x\right)+\mathrm{ln}x\cdot \frac{d}{dx}\left(\mathrm{ln}\left(\mathrm{sin}x\right)\right)$
Thus
$\frac{{y}^{\prime }}{y}=\frac{1}{x}\cdot \mathrm{ln}\left(\mathrm{sin}x\right)+\mathrm{ln}x\cdot \frac{\mathrm{cos}x}{\mathrm{sin}x}$
Simplifying
${y}^{\prime }=y\left(\frac{\mathrm{ln}\left(\mathrm{sin}x\right)}{x}+\mathrm{ln}x\mathrm{cot}x\right)=\left(\mathrm{sin}x{\right)}^{\mathrm{ln}x}\left(\frac{\mathrm{ln}\left(\mathrm{sin}x\right)}{x}+\mathrm{ln}x\mathrm{cot}x\right)$
###### Not exactly what you’re looking for? Taniya Burns
You forgot to use the product rule,
$\frac{d}{dx}\left(\mathrm{ln}x\cdot \mathrm{ln}\left(\mathrm{sin}x\right)\right)=\mathrm{ln}x\cdot \frac{d}{dx}\mathrm{ln}\left(\mathrm{sin}x\right)+\left(\frac{d}{dx}\mathrm{ln}x\right)\cdot \mathrm{ln}\left(\mathrm{sin}x\right)\ne \frac{1}{x}\frac{1}{\mathrm{sin}x}\mathrm{cos}x$