Question

If $f\left( xy \right)=f\left( x \right).f\left( y \right)\forall x,\And f'\left( 1 \right)=2,$then test the differentiability of $f\left( x \right)$.

Answer 1

Hint: Since we have to test the differentiability, we have to find the derivative of the function. We can find derivatives using the first principle of derivative and using the functional relation given in the question.

Complete step-by-step answer:
It is given in the question $f\left( xy \right)=f\left( x \right).f\left( y \right)\text{ }\And f'\left( 1 \right)=2$.
In such types of questions, to test the differentiability of the function$f\left( x \right)$, it is required to find the derivative $f'\left( x \right)$. To find $f'\left( x \right)$, we have to follow certain steps;
1. We will use the first principle to find $f'\left( x \right)$.
According to first principle;
$\begin{aligned} & f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h} \\\ & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x\left( 1+\dfrac{h}{x} \right) \right)-f\left( x \right)}{h}.........\left( I \right) \\\ \end{aligned}$
Since it is given $f\left( xy \right)=f\left( x \right).f\left( y \right)$ we can substitute $f\left( x\left( 1+\dfrac{h}{x} \right) \right)=f\left( x \right).f\left( 1+\dfrac{h}{x} \right)$ in $\left( I \right)$;
$\begin{aligned} & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)f\left( 1+\dfrac{h}{x} \right)-f\left( x \right)}{h} \\\ & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( x \right)\dfrac{\left( f\left( 1+\dfrac{h}{x} \right)-1 \right)}{h} \\\ \end{aligned}$
Since the limit is with respect to $h$, we get functions of $x$ out of the limit.

$\Rightarrow f'\left( x \right)=f\left( x \right)\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-1}{h}.........\left( II \right)$
We cannot proceed further in step 1. So now, we will proceed to step 2.
2. We will find some boundary values of $f\left( x \right)$.
Given $f\left( xy \right)=f\left( x \right).f\left( y \right)$
Let us assume $x=1,y=1$.
$\begin{aligned} & \Rightarrow f\left( 1 \right)=f\left( 1 \right)\times f\left( 1 \right) \\\ & \Rightarrow f\left( 1 \right)={{\left( f\left( 1 \right) \right)}^{2}} \\\ \end{aligned}$
Cancelling $f\left( 1 \right)$ both sides, we get;
$f\left( 1 \right)=1..............\left( III \right)$
From $\left( III \right)$, we will substitute $1$ as $f\left( 1 \right)$ in $\left( II \right)$,
$\Rightarrow f'\left( x \right)=f\left( x \right)\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{h}$
Let us multiple and divide the denominator with $x$.
$\Rightarrow f'\left( x \right)=f\left( x \right)\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{x.\dfrac{h}{x}}$
Since the limit is with respect to $h$, we can take $\dfrac{1}{x}$ out of the limit. Also, we can replace $\underset{h\to 0}{\mathop{\lim }}\,$ by $\underset{\dfrac{h}{x}\to 0}{\mathop{\lim }}\,$.
$f'\left( x \right)=\dfrac{f\left( x \right)}{x}\underset{\dfrac{h}{x}\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{\dfrac{h}{x}}.........\left( IV \right)$
Consider the first principle
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Let us put $x=1,h=\dfrac{h}{x}$.
$\Rightarrow f'\left( 1 \right)=\underset{\dfrac{h}{x}\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{\dfrac{h}{x}}...........\left( V \right)$
So, substituting $\underset{\dfrac{h}{x}\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+\dfrac{h}{x} \right)-f\left( 1 \right)}{\dfrac{h}{x}}$from $\left( V \right)$ in $\left( IV \right)$, we get;
$\Rightarrow f'\left( x \right)=\dfrac{f\left( x \right)}{x}.f'\left( 1 \right)$
It is given in the question $f'\left( 1 \right)=2$.
$\Rightarrow f'\left( x \right)=\dfrac{2f\left( x \right)}{x}..........\left( VI \right)$
Let us substitute $f\left( x \right)=y$
$\Rightarrow f'\left( x \right)=\dfrac{dy}{dx}$
Substituting $f\left( x \right)$and $\text{ }f'\left( x \right)$ in $\left( VI \right)$;
$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{2y}{x} \\\ & \Rightarrow \dfrac{1}{y}dy=2.\dfrac{1}{x}dx \\\ \end{aligned}$
Integrating both sides;
$\int{\dfrac{1}{y}dy=2\int{\dfrac{1}{x}dx}}$
We know $\int{\dfrac{1}{x}dx=lnx}$ and $\int{\dfrac{1}{y}dy=lny}$;

& \Rightarrow lny=2lnx\text{ } \\\ & \Rightarrow lny=ln{{x}^{2}},\text{ }\because 2lnx=ln{{x}^{2}} \\\ & \Rightarrow y={{x}^{2}} \\\ \end{aligned}$$ Differentiating with respect to $x$; $y'=2x$ Let us plot the graph of $y'$;  Since $y'=2x$ is a continuous function, we can say $y$ is differentiable for all $x.$ Sine $y'=2x$, we can say that $f\left( x \right)$ is differentiable $\forall x\in R.$ Note: There is a possibility of mistake while finding the boundary value of the function in step 2. The boundary value is to be found by taking help of the information given in the question. Like it was given in the question that $f'\left( 1 \right)=2$ that is why we had to find the value of $f\left( 1 \right)$ in step 2.