Solution for the question - if f(x+y)=f(x)f(y)+x,y and f(5)=2,f^(')(0)=3 then f^(')(5) is 2610

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Question

If $f left parenthesis x plus y right parenthesis equals f left parenthesis x right parenthesis f left parenthesis y right parenthesis plus x comma y space$ and $f left parenthesis 5 right parenthesis equals 2 comma f to the power of apostrophe left parenthesis 0 right parenthesis equals 3$ then $f to the power of apostrophe left parenthesis 5 right parenthesis$ is

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6
2

hint Hint:

In this question first we will find the value of $f left parenthesis 0 right parenthesis$ by putting $x equals 5 comma y equals 0$ . Then we will take derivative to find the answer.
$f apostrophe left parenthesis x right parenthesis equals limit as h minus greater than 0 of fraction numerator f left parenthesis x plus h right parenthesis minus f left parenthesis h right parenthesis over denominator h end fraction$ .

The correct answer is: 6

In this question we are given $f left parenthesis x plus y right parenthesis equals f left parenthesis x right parenthesis plus f left parenthesis y right parenthesis$ and $f left parenthesis 5 right parenthesis equals 2 comma f apostrophe left parenthesis 0 right parenthesis equals 3$
Step1: Putting $x equals 5 comma y equals 0$ in the given expression
$f left parenthesis 5 plus 0 right parenthesis equals space f left parenthesis 0 right parenthesis f left parenthesis 5 right parenthesis$
=> $f left parenthesis 5 right parenthesis left square bracket f left parenthesis 0 right parenthesis minus 1 right square bracket equals 0$
=> $f left parenthesis 0 right parenthesis equals space 1$
Step2: By taking derivative
We know that $f apostrophe left parenthesis x right parenthesis equals limit as h minus greater than 0 of fraction numerator f left parenthesis x plus h right parenthesis minus f left parenthesis h right parenthesis over denominator h end fraction$
$f apostrophe left parenthesis 5 right parenthesis space equals space limit as h minus greater than 0 of fraction numerator f left parenthesis 5 plus h right parenthesis minus f left parenthesis 5 right parenthesis over denominator h end fraction$
We can write $f left parenthesis 5 plus h right parenthesis space equals space f left parenthesis 5 right parenthesis f left parenthesis h right parenthesis$
$f apostrophe left parenthesis 5 right parenthesis space equals space limit as h minus greater than 0 of fraction numerator f left parenthesis 5 right parenthesis f left parenthesis h right parenthesis minus f left parenthesis 5 right parenthesis over denominator h end fraction$
=> $f apostrophe left parenthesis 5 right parenthesis space equals space limit as h minus greater than 0 of fraction numerator f left parenthesis 5 right parenthesis left parenthesis f left parenthesis h right parenthesis minus 1 right parenthesis over denominator h end fraction$
By substituting $1 space equals space f left parenthesis 0 right parenthesis$
=> $f apostrophe left parenthesis 5 right parenthesis space equals space limit as h minus greater than 0 of fraction numerator f left parenthesis 5 right parenthesis left parenthesis f left parenthesis h right parenthesis minus f left parenthesis 0 right parenthesis right parenthesis over denominator h end fraction$
=> $f apostrophe left parenthesis 5 right parenthesis space equals space space f left parenthesis 5 right parenthesis limit as h minus greater than 0 of fraction numerator left parenthesis f left parenthesis h right parenthesis minus f left parenthesis 0 right parenthesis right parenthesis over denominator h end fraction$
=> $f apostrophe left parenthesis 5 right parenthesis space equals space space f left parenthesis 5 right parenthesis f apostrophe left parenthesis 0 right parenthesis$
=> $f apostrophe left parenthesis 5 right parenthesis equals 2 cross times 3 equals 6$

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The correct answer is: 6

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If and then is

0162

In this question first we will find the value of by putting . Then we will take derivative to find the answer..

The correct answer is: 6

In this question we are given and Step1: Putting in the given expression=> =>Step2: By taking derivativeWe know that We can write => By substituting => =>=> =>

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For $x element of R L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x minus 3 right parenthesis divided by left parenthesis x plus 2 right parenthesis right parenthesis to the power of x equals$

For $x element of R L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x minus 3 right parenthesis divided by left parenthesis x plus 2 right parenthesis right parenthesis to the power of x equals$