# How to find the derivative of $f(x)$?

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

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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