Question

Suppose that for all $x,y\in R$, $f\left( x+y \right)=f\left( x \right).f\left( y \right)$ and $f'\left( 0 \right)$ exists. Then show that $f'\left( x \right)$ exists and equals to $f\left( x \right).f'\left( 0 \right)$ for all $x\in R$.

Answer 1

To solve this question, we will use the definition of differentiability of a function and we will simplify it by using some properties of limit in the process of the solution of this question. Then, the solution will lead to the result that we need to verify.

Complete step by step solution:
Since, we have from the question as:
$\Rightarrow f\left( x+y \right)=f\left( x \right).f\left( y \right)$ and $f'\left( 0 \right)$ exists for all $x,y\in R$.
Now, we will use the definition of differentiability of a function when we have $x\in R$ and $h\ne 0$. We will write it as:
$\Rightarrow f'\left( x \right)=\underset{ h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Here, we will use the given statement $f\left( x+y \right)=f\left( x \right).f\left( y \right)$ in the above step to simplify it to proceed further as:
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)\cdot f\left( h \right)-f\left( x \right)}{h}$
Now, we can take common factor in the numerator as:
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)\cdot \left( f\left( h \right)-1 \right)}{h}$
Here, we will use the property of product of two limit functions as:
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( x \right)\cdot \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( f\left( h \right)-1 \right)}{h}$
Since, $h$ is constant and tends to $0$ and from the definition of differentiability of a function, it will be $f'\left( 0 \right)$ and from the given statement it exists. So, we can write the above step as:
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( x \right)\cdot f'\left( 0 \right)$
Here, we will use the property of limit on a constant. Then the above step will be as:
$\Rightarrow f'\left( x \right)=f\left( x \right)\cdot f'\left( 0 \right)$
This we need to verify. Hence, this completes the proof.

Note: A function is differentiable at any point if and only if it has a limit or it has a derivative. Similarly, if the limit of a function does not exist, the function would not be differentiable. We can define it in terms of mathematics as:
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Where, $x\in R$ and $h\ne 0$.