Question

If $f\left( {x + y} \right) = f\left( x \right)f\left( y \right)$ for all $x,y \in R$, $f\left( 5 \right) = 2$, and $f'\left( 0 \right) = 3$, then $f'\left( 5 \right)$ equals
(a) 6
(b) 3
(c) 5
(d) None of these

Answer 1

Here, we need to find the value of $f'\left( 5 \right)$. First, we will find the value of $f\left( 0 \right)$. Then, we will differentiate the given function and simplify the equation. Finally, we will substitute the values in the derivative and, using the given information, find the required value of $f'\left( 5 \right)$.
Formula Used: The product rule of differentiation states that the derivative of the product of two functions is given as $\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{d\left( v \right)}}{{dx}} + \dfrac{{d\left( u \right)}}{{dx}}v$.

Complete step by step solution:
First, we will find the value of $f\left( 0 \right)$.
Substituting $x = 5$ and $y = 0$ in the equation $f\left( {x + y} \right) = f\left( x \right)f\left( y \right)$, we get
\Rightarrow f\left( {5 + 0} \right) = f\left( 5 \right)f\left( 0 \right) \\\ \Rightarrow f\left( 5 \right) = f\left( 5 \right)f\left( 0 \right) \\\
Substituting $f\left( 5 \right) = 2$ in the expression, we get
\Rightarrow 2 = 2 \times f\left( 0 \right) \\\ \Rightarrow 2 = 2f\left( 0 \right) \\\
Dividing both sides of the equation by 2, we get
$\Rightarrow \dfrac{2}{2} = \dfrac{{2f\left( 0 \right)}}{2}$
Thus, we get
$\Rightarrow f\left( 0 \right) = 1$
Now, we will differentiate the given function.
Differentiating both sides of the equation $f\left( {x + y} \right) = f\left( x \right)f\left( y \right)$ with respect to $x$, we get
$\dfrac{{d\left[ {f\left( {x + y} \right)} \right]}}{{dx}} = \dfrac{{d\left[ {f\left( x \right)f\left( y \right)} \right]}}{{dx}}$
The product rule of differentiation states that the derivative of the product of two functions is given as $\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{d\left( v \right)}}{{dx}} + \dfrac{{d\left( u \right)}}{{dx}}v$.
Simplifying the equation using the product rule of differentiation, we get
$\Rightarrow \dfrac{{d\left[ {f\left( {x + y} \right)} \right]}}{{dx}} = f\left( x \right)\dfrac{{d\left[ {f\left( y \right)} \right]}}{{dx}} + \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}f\left( y \right)$
Simplifying the equation using the chain rule of differentiation, we get
$\Rightarrow f'\left( {x + y} \right)\dfrac{{d\left( {x + y} \right)}}{{dx}} = f\left( x \right)f'\left( y \right)\dfrac{{d\left( y \right)}}{{dx}} + f'\left( x \right)f\left( y \right)$
The differential $\dfrac{{d\left( {x + y} \right)}}{{dx}}$ is the derivative of the sum of two functions.
Therefore, we get
$\Rightarrow f'\left( {x + y} \right)\left[ {\dfrac{{d\left( x \right)}}{{dx}} + \dfrac{{d\left( y \right)}}{{dx}}} \right] = f\left( x \right)f'\left( y \right)\dfrac{{d\left( y \right)}}{{dx}} + f'\left( x \right)f\left( y \right)$
Simplifying the expression and rewriting $\dfrac{{d\left( y \right)}}{{dx}}$ as $y'$, we get
$\Rightarrow f'\left( {x + y} \right)\left[ {1 + y'} \right] = f\left( x \right)f'\left( y \right)y' + f'\left( x \right)f\left( y \right)$
Substituting $x = 5$ and $y = 0$ in the equation, we get
\Rightarrow f'\left( {5 + 0} \right)\left[ {1 + y'} \right] = f\left( 5 \right)f'\left( 0 \right)y' + f'\left( 5 \right)f\left( 0 \right) \\\ \Rightarrow f'\left( 5 \right)\left[ {1 + y'} \right] = f\left( 5 \right)f'\left( 0 \right)y' + f'\left( 5 \right)f\left( 0 \right) \\\
Substituting $f\left( 0 \right) = 1$, $f\left( 5 \right) = 2$, and $f'\left( 0 \right) = 3$ in the equation, we get
$\Rightarrow f'\left( 5 \right)\left[ {1 + y'} \right] = 2 \times 3 \times y' + f'\left( 5 \right) \times 1$
Multiplying the terms in the expression, we get
$\Rightarrow f'\left( 5 \right)\left[ {1 + y'} \right] = 6y' + f'\left( 5 \right)$
Multiplying $f'\left( 5 \right)$ by $\left[ {1 + y'} \right]$ using the distributive law of multiplication, we get
$\Rightarrow f'\left( 5 \right) + f'\left( 5 \right)y' = 6y' + f'\left( 5 \right)$
Subtracting $f'\left( 5 \right)$ from both sides of the equation, we get
\Rightarrow f'\left( 5 \right) + f'\left( 5 \right)y' - f'\left( 5 \right) = 6y' + f'\left( 5 \right) - f'\left( 5 \right) \\\ \Rightarrow f'\left( 5 \right)y' = 6y' \\\
Dividing both sides of the equation by $y'$, we get
$\Rightarrow \dfrac{{f'\left( 5 \right)y'}}{{y'}} = \dfrac{{6y'}}{{y'}}$
Thus, we get
$\therefore f'\left( 5 \right) = 6$
Therefore, we get the value of $f'\left( 5 \right)$ as 6.

Thus, the correct option is option (a).

Note:
We have used the distributive law of multiplication in the solution. The distributive law of multiplication states that $a\left( {b + c} \right) = a \cdot b + a \cdot c$.
We rewrote the differential $\dfrac{{d\left( {x + y} \right)}}{{dx}}$ as$\dfrac{{d\left( x \right)}}{{dx}} + \dfrac{{d\left( y \right)}}{{dx}}$ because $\dfrac{{d\left( {x + y} \right)}}{{dx}}$ is the derivative of the sum of two functions. The derivative of the sum of two functions is the sum of the derivatives of the two functions, that is $\dfrac{{d\left[ {f\left( x \right) + g\left( x \right)} \right]}}{{dx}} = \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} + \dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}$.