Answer
Verified
481.2k+ 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,
\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}\]
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,
\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}\]
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Change the following sentences into negative and interrogative class 10 english CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Application to your principal for the character ce class 8 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE