Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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\[f'(2)\].

seo-qna
Last updated date: 14th Jul 2024
Total views: 451.5k
Views today: 12.51k
Answer
VerifiedVerified
451.5k+ views
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,
\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}\]