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

Hint:
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 \[\dfrac{{d\left[ {f\left( x \right) + g\left( x \right)} \right]}}{{dx}} = \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} + \dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}\].