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# Let, $f(xy)=f(x)f(y)\forall x,y\ in R$ and $f$ is differentiable at $x=1$ such that $f'(1)=1$.Also, $f(1)\ne 1,f(2)=3$. Then find$f'(2)$.

Last updated date: 14th Jul 2024
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Hint: Use the concept derivative using first principle to solve the above problem i.e. use the formula $f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$.

We will write the given values first,
$f(1)\ne 0,f'(1)=1,f(2)=3$………………………………………… (1)
Now we will write the given equation,
$f(xy)=f(x)f(y)$…………………………………………………… (2)
Put, $x=y=1$so that we can calculate the value of f (1),
$\therefore f(1\times 1)=f(1)\times f(1)$
$\therefore f(1)-f(1)\times f(1)=0$
$\therefore f(1)\times [1-f(1)]=0$
$\therefore f(1)=0$ OR $[1-f(1)]=0$
We have given $f(1)\ne 0$,
$\therefore [1-f(1)]=0$
$\therefore f(1)=1$…………………………………………………………………… (3)
As we have to find $f'(2)$ we should know the formula for differentiation by First Principle.

Formula:
$f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$
By using above formula we can write equation for $f'(2)$,
$\therefore f'(2)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(2+h)-f(2)}{h}$
Now take 2 common from $f(2+h)$,
$\therefore f'(2)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left[ 2(1+\dfrac{h}{2}) \right]-f(2)}{h}$…………………………………….. (4)
Now using the equation given in problem,
$f(xy)=f(x)\times f(y)$
We can write $f\left[ 2\left( 1+\dfrac{h}{2} \right) \right]$ by comparing it with above equation,
$\therefore f\left[ 2\left( 1+\dfrac{h}{2} \right) \right]=f(2)\times f\left( 1+\dfrac{h}{2} \right)$
Now put this value in equation (4)
$\therefore f'(2)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(2)\times f\left( 1+\dfrac{h}{2} \right)-f(2)}{h}$
By taking $f(2)$ common from the equation we can write,
$\therefore f'(2)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(2)\times \left[ f\left( 1+\dfrac{h}{2} \right)-1 \right]}{h}$
As $f(2)$ is constant we can take it outside the limit,
$\therefore f'(2)=\underset{h\to 0}{\mathop{f(2)\lim }}\,\dfrac{\left[ f\left( 1+\dfrac{h}{2} \right)-1 \right]}{h}$

Now 1 can be replaced by $f(1)$ as we have evaluated it in equation (3),
$\therefore f'(2)=\underset{h\to 0}{\mathop{f(2)\lim }}\,\dfrac{\left[ f\left( 1+\dfrac{h}{2} \right)-f(1) \right]}{h}$
If we observe above equation carefully then we can compare the equation with the formula of first principle,
$f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$
After comparing we can conclude that the equation requires adjustment in the denominator, i.e. we have to replace ‘h’ by $\dfrac{h}{2}$.
For that we will do the adjustment of multiplying and dividing the equation by $\dfrac{1}{2}$,
$\therefore f'(2)=f(2)\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left[ f\left( 1+\dfrac{h}{2} \right)-f(1) \right]\times \dfrac{1}{2}}{h\times \dfrac{1}{2}}$………………………………………. (5)
By using formula of first principle we can write the equation for $f'(1)$ by replacing ‘h’ by $\dfrac{h}{2}$ as follows,

$f'(1)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(1+\dfrac{h}{2})-f(1)}{\dfrac{h}{2}}$
Therefore equation (5) becomes,
$\therefore f'(2)=f(2)\times f'(1)\times \dfrac{1}{2}$
Put the values of equation (1) in above equation,
$\therefore f'(2)=3\times 1\times \dfrac{1}{2}$
$\therefore f'(2)=\dfrac{3}{2}$
Therefore the value of $f'(2)$ is $\dfrac{3}{2}$.
Note: You can try to solve this type of problems by using first principles if you don’t have any idea.
First principle of derivative can be given as,
$f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$