Question

Let $f(x + y) = f(x)f(y)$ for all $x,y \in R$ and suppose that f is differentiable at 0 and $f'(0) = 4$. If $f({x_0}) = 8$ then $f'({x_0})$ is equal to

Answer 1

Hint: We’ve been given the equation of the function f and some value related to the function. Now we will substitute ‘y’ with ${x_0}$ as we have given the value of the function at ${x_0}$, now on differentiating we’ll get another equation where we will substitute ‘x’ with zero as the input of the derivative function will become ${x_0}$ and that is what is asked i.e. the derivative of the function at ${x_0}$ and also we have given the value of the derivative function at zero, so we will get the required value.
Now solving this equation we will get the required answer.

Complete step by step answer:

Given data: $f(x + y) = f(x)f(y)$
$f'(0) = 4$
$f({x_0}) = 8$
Now solving for $f(x + y) = f(x)f(y)$
On substituting $y = {x_0}$
$ \Rightarrow f(x + {x_0}) = f(x)f({x_0})$
Now, substituting $f({x_0}) = 8$
$ \Rightarrow f(x + {x_0}) = f(x)8$
Now, differentiating with-respect-to x
$ \Rightarrow f'(x + {x_0}) = f'(x)8$
Now substituting the value of ‘x’ i.e. $x = 0$
We have,
$ \Rightarrow f'(0 + {x_0}) = f'(0)8$
Now, substituting the value of $f'(0)$ as $f'(0) = 4$
$ \Rightarrow f'({x_0}) = 4(8)$
On simplifying we have
$\therefore f'({x_0}) = 32$
Therefore the required value is 32.

Note: Some of the students differentiate the given equation of function f but as the function has two independent variables i.e. ‘x’ and ‘y’ and we have no relation in between ‘y’ and ‘x’ to get that derivative of one with-respect-to other independent variables so even if we try of eliminating it, that will take much more of the process to get to the required answer, so we used the method that is applied for simple simplification.