Question

If $f(x)=\log x$, then show that f is differentiable at $f\in \left( 0,\infty \right)$ and $f'(x)=\dfrac{1}{x}$.

Answer 1

We check if there are points on which the function $f\left( x \right)=\log x$ is not defined. We use the information that a function is differentiable at $x=a$ if it is continuous at $x=a$ and then find derivative at $x=a$ is given by the expression $\displaystyle \lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}$ where $h$ is an infinitesimally small positive quantity.

Complete step-by-step solution
We have been given that $f(x)=\log x$. We know that the function $f\left( x \right)=\log x$ is not defined for negative numbers and 0 so the given function is well defined.
Now, for $x>0$, we know that the given function $f(x)=\log x$ is continuous and differentiable.
Function $f(x)$ is said to be differentiable at the point $x=a$ if the derivative of the function exists at every point in the given domain. $$$$
Now, we know that the differentiability formula is given with infinitesimally small positive quantity by
$f'(a)=\displaystyle \lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}$
So, the differentiation of given function $f(x)=\log x$ with respect to $x$ will be with derivative at all points
$f'(x)=\displaystyle \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$
Now, we put the value$f(x)=\log x$, we get
$\Rightarrow f'(x)=\displaystyle \lim_{h\to 0}\dfrac{\log (x+h)-\log x}{h}$
Now, we know the logarithmic identity $\log m-\log n=\log \dfrac{m}{n}$ So, we have

& \Rightarrow f'(x)=\displaystyle \lim_{h\to 0}\dfrac{\log \dfrac{\left( x+h \right)}{x}}{h} \\\ & \Rightarrow f'(x)=\displaystyle \lim_{h\to 0} \dfrac{\log \left( 1+\dfrac{h}{x} \right)}{h} \\\ \end{aligned}$$ Let us multiply $\dfrac{1}{x}$ in the numerator and denominator in the bracket of the logarithm and have, $$\Rightarrow f'(x)=\displaystyle \lim_{h\to 0} \dfrac{\log\left( 1+\dfrac{h}{x} \right)}{\dfrac{h}{x}}\times \dfrac{1}{x}$$ We use the standard limit $\displaystyle \lim_{x \to 0}\dfrac{\log (1+x)}{x}=1$ for $x=\dfrac{h}{x}$ in the above step to have, $$\begin{aligned} & \Rightarrow f'(x)=\displaystyle \lim_{h\to 0}1\times \dfrac{1}{x} \\\ & \Rightarrow f'(x)=\dfrac{1}{x} \\\ \end{aligned}$$ Hence proved that $f'(x)=\dfrac{1}{x}$. **Note:** Function $f(x)$ is said to be differentiable at the point $x=a$ if the derivative of the function exists at every point in the given domain. $f'(x)$ is the first order derivative of the function. To solve this type of question one must remember the formula of differentiability. The limit of the function at $x=a$ exists if the left hand limit and right hand limit is equal. The function is continuous at the point $x=a$ if limit exists and is equal to the functional value at $x=a$ that is $f\left( a \right)=\displaystyle \lim_{x \to a}f\left( x \right)$.