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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,

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}\]

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,

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}\]

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