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Is the question all right??
Put y=0 in the given functional equation to get:
$0+2=f(x)+0+0~~~~\small{\text{(since f(0)=0)}}$$\Large f(x)=2$
which does not satisfy the condition f'(1)=2 !!

(It might work if indeed it was given f(1)=2 and f'(0)=0)

I also came up with the same conclusion as you, i.e. $f(x)=2$, so i thought of using the $f'(1)=2$.
So as we know from the definition of derivatives

$f'(1)=\frac { f(1+h)-f(1) }{ h }\\$

$\Longrightarrow f'(1)=\frac { f(1(1+h))-f(1) }{ h }$

From the functional equation we have,$\\$

$f(1)f(1+h))+2=f(1)+f(1+h)+f(1(1+h))$

$\Longrightarrow f(1+h)=f(1)f(1+h)-f(1)-f(1+h)+2$

$\therefore \quad f^{ 1 }\left( x \right) =\frac { f(1)f(1+h)-f(1)-f(1+h)+2-f(1) }{ h } =\frac { f(1)(f(1+h)-2)-(f(1+h)-2) }{ h }$

$\Longrightarrow f^{ 1 }\left( x \right) =\frac { (f(1)-1)(f(1+h)-2) }{ h }$

$\Longrightarrow 2h=(f(1)-1)(f(1+h)-2)$

From here on I am not able to conclude anything useful, so if you would help me lead in the right direction or tell me something that i am doing wrong.

But your suggestion makes the functional equation pretty much useless, as for $x=1$, and $y=c$, $c\in dom(f)$ we do not get any relation to solve for $f(x)$

Hey, if you see the other post, which is the same question, here,
you'll see we get, $f(x) - 1 = x^t$
For no value of t, will both the given conditions be satisfied- there is no such function, as a result. Try it out! I think you should check your question again.

The question is wrong . I will prove it .
Put $x,y=0$$f(0)^{2}-3f(0)+2=0$
This implies $f(0)=1,2$
now put $x,y=1$
this gives $f(1)=1,2$
Differentiating the function partially with respect to $x$ we get
$f'(x)f(y)=f'(x)+yf'(xy).......(1)$
Now put $x,y=1$ in $(1)$
We get $f(1)=2$
Now put $x,y=0$ in $(1)$
We get$f(0)=1..or..f'(0)=0$
Now put $x=1$ in (1)
we get $\frac{2f(y)-2}{y}=f'(y)...(2)$
Now let $f(0)=1$ , we have $f(1)=2$then we observe that $f(x)=x+1$
also in satisfies main functional equation and (2),(1) but $f'(x)=1$ hence question is wrong .
If only main functional equation is given without any condition then $f(x)=x+1$.

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## Comments

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TopNewestIs the question all right?? Put y=0 in the given functional equation to get: $0+2=f(x)+0+0~~~~\small{\text{(since f(0)=0)}}$ $\Large f(x)=2$ which does not satisfy the condition f'(1)=2 !!

(It might work if indeed it was given f(1)=2 and f'(0)=0)

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I also came up with the same conclusion as you, i.e. $f(x)=2$, so i thought of using the $f'(1)=2$. So as we know from the definition of derivatives

$f'(1)=\frac { f(1+h)-f(1) }{ h }\\$

$\Longrightarrow f'(1)=\frac { f(1(1+h))-f(1) }{ h }$

From the functional equation we have,$\\$

$f(1)f(1+h))+2=f(1)+f(1+h)+f(1(1+h))$

$\Longrightarrow f(1+h)=f(1)f(1+h)-f(1)-f(1+h)+2$

$\therefore \quad f^{ 1 }\left( x \right) =\frac { f(1)f(1+h)-f(1)-f(1+h)+2-f(1) }{ h } =\frac { f(1)(f(1+h)-2)-(f(1+h)-2) }{ h }$

$\Longrightarrow f^{ 1 }\left( x \right) =\frac { (f(1)-1)(f(1+h)-2) }{ h }$

$\Longrightarrow 2h=(f(1)-1)(f(1+h)-2)$

From here on I am not able to conclude anything useful, so if you would help me lead in the right direction or tell me something that i am doing wrong.

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You have to take the limit for the derivative, as h goes to 0.

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Please if you can post a solution@parv mor

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f(1)=2 and f'(0)=0 should be there for the question to be correct

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But your suggestion makes the functional equation pretty much useless, as for $x=1$, and $y=c$, $c\in dom(f)$ we do not get any relation to solve for $f(x)$

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Ya.. I also reached the same conclusion...

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Hey, if you see the other post, which is the same question, here,

you'll see we get,

$f(x) - 1 = x^t$

For no value of t, will both the given conditions be satisfied- there is no such function, as a result. Try it out! I think you should check your question again.

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The question is wrong . I will prove it . Put $x,y=0$ $f(0)^{2}-3f(0)+2=0$ This implies $f(0)=1,2$ now put $x,y=1$ this gives $f(1)=1,2$ Differentiating the function partially with respect to $x$ we get $f'(x)f(y)=f'(x)+yf'(xy).......(1)$ Now put $x,y=1$ in $(1)$ We get $f(1)=2$ Now put $x,y=0$ in $(1)$ We get$f(0)=1..or..f'(0)=0$ Now put $x=1$ in (1) we get $\frac{2f(y)-2}{y}=f'(y)...(2)$ Now let $f(0)=1$ , we have $f(1)=2$then we observe that $f(x)=x+1$ also in satisfies main functional equation and (2),(1) but $f'(x)=1$ hence question is wrong . If only main functional equation is given without any condition then $f(x)=x+1$.

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