Let a function $$f:\Re \to \Re $$ be given by $$f(x + y) = f(x)f(y)$$ for all$$x,y \in \Re $$. If the function$$f(x)$$ is differentiable at$$x = 0$$, to show that $$f'(x) = f'(0)f(x)$$ for all$$x \in \Re $$and also to determine the value of$$f(x)$$.

Question

Let a function $$f:\Re  \to \Re $$ be given by $$f(x + y) = f(x)f(y)$$ for all$$x,y \in \Re $$. If the function$$f(x)$$ is differentiable at$$x = 0$$, to show that $$f'(x) = f'(0)f(x)$$ for all$$x \in \Re $$and also to determine the value of$$f(x)$$.

Vedantu Content Team · Accepted Answer

Hint: we have already known that If$$f:\Re  \to \Re $$ is a functionThen, $$(f(x)f(y))' = f'(x)f(y) + f(x)f'(y)$$Where, $$f(x)$$ is differentiable on$$\Re $$.Complete step-by-step answer:Let us consider the given function $$f:\Re  \to \Re $$ which is given by $$f(x + y) = f(x)f(y)$$ for all $$x,y \in \Re $$. It is also given that the function$$f(x)$$ is differentiable at $$x = 0$$Here we have that is, $$f(x + y) = f(x)f(y)$$ in given.Now we are going to differentiate the given function for$$x$$, we get,$$f'(x + y) = (f(x)f(y))'$$From the given hint we can substitute the value of$$(f(x)f(y))'$$,Hence now we get after the substitution of the value of$$(f(x)f(y))'$$,$$f'(x + y) = f'(x)f(y) + f(x)f'(y)$$We have to find the value of the function $$x = 0$$, we will substitute the $$x$$by zero.So substitute $$x = 0$$ in the function we get,$$f'(0 + y) = f'(0)f(y) + f(0)f'(y)$$Here, $$f(y)$$is a function of$$y$$. So, $$f'(y) = 0$$ for any values of$$x$$.Now we are going to substitute $$f'(y) = 0$$ in the above-given step we get,$$f'(y) = f'(0)f(y)$$Since $$y$$ is a variable, we can change the variable $$y$$ by a new variable $$x$$ since$$x,y \in \Re $$. If the function$$f(x)$$ is differentiable at$$x = 0$$,$$f'(x) = f'(0)f(x)$$for all $$x \in \Re $$Hence we have shown that$$f'(x) = f'(0)f(x)$$for all $$x \in \Re $$.As we have proved the above function we know that,$$f'(x) = f'(0)f(x)$$Now we have to find$$f(x)$$ of the function we can divide the above equation by$$f'(0)$$on both sides,On dividing the term we get,$$f(x) = \dfrac{{f'(x)}}{{f'(0)}}$$, provided $$f'(0) \ne 0$$Now we have determined that $$f(x) = \dfrac{{f'(x)}}{{f'(0)}}$$for all $$x \in \Re $$.Hence, we have shown that $$f'(x) = f'(0)f(x)$$ for all $$x \in \Re $$And,$$f(x) = \dfrac{{f'(x)}}{{f'(0)}}$$, for all $$x \in \Re $$ provided $$f'(0) \ne 0$$Note:If $$f'(0) = 0$$then the determined functional value of $$f(x)$$ will become undefined. Hence not to get an undefined value for$$f(x)$$the differentiation of the function at $$x = 0$$has to be a non-zero value.

Let a function \[f:\Re \to \Re \] be given by \[f(x + y) = f(x)f(y)\] for all\[x,y \in \Re \]. If the function\[f(x)\] is differentiable at\[x = 0\], to show that \[f'(x) = f'(0)f(x)\] for all\[x \in \Re \]and also to determine the value of\[f(x)\].