Proof: Derivative of ln(x) = 1/x by First Principles
Proof: Derivative of ln(x) = 1/x by First Principles

Differentiate the Function $f(x)= \sqrt{x} \ln x$

• $\begingroup$ Confirm how you took logs of both sides of $y = \sqrt{x}\ln x$? $\endgroup$– Chinny84Jun 30, 2015 at 15:56
• $\begingroup$ Yes. I am assuming I have to differentiate before I take the log? $\endgroup$ Jun 30, 2015 at 16:03

$f(x) = \sqrt{x} \ln x$. Using the product rule we have that $$f'(x) = \ln x \cdot \frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x} + \sqrt{x}\cdot\frac{\mathrm{d}}{\mathrm{d}x} \ln x$$

Hence $$f'(x) = \frac{\ln x}{2\sqrt{x}} + \frac{\sqrt{x}}{x}.$$

Further simplification results in $$f'(x) = \frac{\ln x}{2\sqrt{x}} + \frac{2}{2\sqrt{x}} = \frac{\ln x + 2}{2\sqrt{x}}$$

Doing it your way: $$\ln f(x) = \ln\left(\sqrt{x} \ln x\right) = \frac{1}{2}\ln x + \ln (\ln x).$$

Then differentiating both sides with respect to $x$ yields $$\frac{f'(x)}{f(x)} = \frac{1}{2x} + \frac{1}{x\ln x}.$$

Multiplying both sides by $f(x)$ gives

$$f'(x) = \sqrt{x} \ln x \left(\frac{1}{2x} + \frac{1}{x \ln x}\right) = \frac{\ln x + 2}{2\sqrt{x}}$$

• $\begingroup$ So although there is ln on the right side already, I need to place another by the term which lacks one? $\endgroup$ Jun 30, 2015 at 16:06
• 2$\begingroup$ If you have $f(x) = \sqrt{x} \ln x$ then taking the logs of both sides means that you do $\ln f(x) = \ln (\sqrt{x} \ln x)$ which you can then split up by using the product rule for logs. Basically, if you $\ln$ one side, then you need to $\ln$ the other side as well! $\endgroup$ Jun 30, 2015 at 16:07

Why did you not take the derivative of $\sqrt{x}$ in step 2? Also, verify what it means to take the log of both sides of an equation. If you want to simply fix your solution, those are my suggestions.
Alternatively, I propose this could be solved with the chain rule. $$f'(x)=\frac{1}{2\sqrt{x}}\ln(x)+\sqrt{x}\frac{1}{x}=\frac{\ln(x)}{2\sqrt{x}}+\frac{2}{2\sqrt{x}}$$

It seems you want to use the logarithmic derivative: $$\ln f(x)=\ln(\sqrt{x}\ln x)=\ln\sqrt{x}+\ln\ln x= \frac{1}{2}\ln x+\ln\ln x$$ Then, differentiating both sides, $$\frac{f'(x)}{f(x)}=\frac{1}{2}\frac{1}{x}+\frac{1}{\ln x}\frac{1}{x}= \frac{\ln x+2}{2x\ln x}$$ and therefore $$f'(x)=\frac{\ln x+2}{2x\ln x}\sqrt{x}\ln x=\frac{\ln x+2}{2\sqrt{x}}$$ Strictly speaking, this works only for $\sqrt{x}\ln x>0$, but it can be shown the method is sound also for the other case.

You have several mistakes. First, the derivative of $$\sqrt{x}\ln x$$ is not $$\sqrt{x}\frac{1}{x}$$ because you have to apply the product rule. But the main error is in the previous step, where you take the logarithm of one side but not also of the other side.

How do you go from $y=\ln (x)\sqrt{x}$ to $\ln y=\ln (x)\sqrt{x}$?

This should be

$$\ln y= \ln(\ln (x)\sqrt{x})=\ln(\ln (x))+0.5\ln(x)$$

As others suggested, you can just use the product rule.

Solution 1 (logarithmic deriviative):

$$y=\sqrt{x}\ln{(x)}$$ $$\ln(y)=\ln{(\sqrt{x}\ln{(x)})}$$ $$\frac{y’}{y}=\ln{(\sqrt{x})}+\ln{(\ln{(x)})}$$ $$\frac{y’}{y}=\frac{1}{2x}+\frac{1}{x\ln{(x)}}$$ $$y’=\frac{\sqrt{x}\ln{(x)}}{2x}+\frac{\sqrt{x}\ln{(x)}}{x\ln{(x)}}$$ $$y’=\frac{2+\ln{(x)}}{2\sqrt{x}}$$

Solution 2 (product rule):

$$y=\sqrt{x}\ln{(x)}$$ $$y’=\frac{\ln(x)}{2\sqrt{x}}+\frac{1}{\sqrt{x}}$$ $$y’=\frac{2+\ln{(x)}}{2\sqrt{x}}$$

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