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This is not a homework problem.

How to find $f'(x)$ if $f(x)=2x^{1/2}\log(2)$?

Thanks in advance for your help! I just can't figure it out...

What rule should I use?

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$f(x)=(2\log 2)x^{1/2}$. Note that $2\log 2$ is a constant, and that the derivative of $x^{1/2}$ is $(1/2)x^{-1/2}$. – André Nicolas Dec 24 '11 at 23:56
up vote 1 down vote accepted

You can use simple differentiation but another way to do this is to find the natural logarithm of both sides and use the fact that the derivative of $\ln f(x)$ is $\frac{f'(x)}{f(x)}$, where $f'(x)$ is the derivative of $f(x)$ Thus, $$\ln f(x)=\ln 2+\frac{1}{2}\ln x+\ln (log(2))$$

So differentiating, you get $$\frac{f'(x)}{f(x)}=\frac{1}{2x}$$ , since $ln 2$ and $ln (\log 2)$ are constants. So you get;

$$f'(x)=\frac{1}{2x}f(x)=x^{-\frac{1}{2}}\log (2)$$

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This answer is undoubtedly correct, but I don't think it's necessary to invoke the product rule or logarithmic differentiation to differentiate $x^{n}$ (multiplied by a constant). Here $n$ is $\frac12$. – Srivatsan Dec 25 '11 at 0:14
You are absolutely correct. That is why I said this is another way. There are several ways like the one posted as a comment under the question. I can delete it, but it might be useful in the future. – smanoos Dec 25 '11 at 0:17
Well, as long as we're looking for other ways to do the problem, square both sides, getting $f^2=C^2x$ where $C$ is the constant $2\log2$, differentiate to get $2ff'=C^2$, so $f'=C^2/2f=C^2/(2C\sqrt x)=C/(2\sqrt x)$, etc. – Gerry Myerson Dec 25 '11 at 1:27

Note that $f(x)=(2\log 2)x^{1/2}$, and that $2\log 2$ is a constant. The derivative of $x^{\frac{1}{2}}$ with respect to $x$ is $\left(\frac{1}{2}\right)x^{\frac{1}{2}-1}$, and therefore $$f'(x)=(2\log 2)\frac{1}{2}x^{-1/2}=(\log 2)x^{-1/2}.$$

Comment: If $k$ is a constant, and $f(x)=kg(x)$, then $f'(x)=kg'(x)$. In more fancy language, if $g'(x)$ exists then $f'(x)$ exists and is equal to $kg'(x)$. This rule should be after a while automatic. In general, if $a$ and $b$ are constants, and $f(x)=ag(x)+b$, then $f'(x)=ag'(x)$. The other rule that we used is the one that says that if $k$ is a constant, then the derivative of $x^k$ is $kx^{k-1}$. This one is usually proved first for $k$ a positive integer, then for $k$ an integer, then for $k$ a rational, and usually somewhat later for all real $k$.

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Note that $\frac{d}{dx} (c \cdot f(x)) = c \cdot f'(x)$. Hence you just want to find the derivative of $f(x) = \sqrt{x}$. Using the definition of the derivative we have

\begin{align*} f'(x) & = \lim_{h \to 0} \: \frac{f(x+h)- f(x)}{h} \\ &= \lim_{h \to 0} \: \frac{\sqrt{x+h} - \sqrt{x}}{h} \\ &= \lim_{h \to 0} \: \frac{\sqrt{x+h}-\sqrt{x}}{h} \times \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\ &=\lim_{h \to 0} \: \frac{x+h - x}{h \cdot \bigl(\sqrt{x+h} + \sqrt{x}\:\bigr)} \\ &= \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2 \cdot \sqrt{x}} \end{align*}

Use the above rule and see that $ \displaystyle f'(x) = 2 \cdot \log(2) \cdot \frac{1}{2 \cdot \sqrt{x}}$

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