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If f(x)=(f(x))/y+(f(y))/x holds for all ...

If `f(x)=(f(x))/y+(f(y))/x` holds for all real `x` and `y` greater than `0a n df(x)` is a differentiable function for all `x >0` such that `f(e)=1/e ,t h e nfin df(x)dot`

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We have
`f(xy)=(f(x))/(y)+(f(y))/(x)" for all "x, y gt0`
`"or "f(1)=f(1)+f(1)" [Putting x=y=1]"`
`"or "f(1)=0`
Now, f(x) is differentiable for all x `gt`0. Therefore,
`f'(x)=underset(hrarr0)lim(f(x+h)-f(x))/(h)`
`=underset(hrarr0)lim(f{x(1+(h)/(x))}-f(x))/(h)`
`=underset(hrarr0)lim((f(x))/(1+(h)/(x))+(f(1+(h)/(x)))/(x)-f(x))/(h)`
`=underset(hrarr0)lim(((-h)/(x)f(x))/(1+(h)/(x))+(f(1+(h)/(x)))/(x))/(h)`
`=-(f(x))/(x)+underset(hrarr0)lim(f(1+(h)/(x)))/(hx)`
`=-(f(x))/(x)+(1)/(x^(2))underset(hrarr0)lim(f(1+(h)/(x)))/((h)/(x))`
`=(-f(x))/(x)+(A)/(x^(2))," where "A=underset(hrarr0)lim(f(1+(h)/(x)))/((h)/(x))`
`therefore" "(d)/(dx)(f(x))=-(f(x))/(x)=(A)/(x^(2))`
`"or "(d)/(dx)(f(x))+(f(x))/(x)=(A)/(x^(2))`
`"or "x(d)/(dx)(f(x))+f(x)=(A)/(x)`
`"or "(d)/(dx)[xf(x)]=(A)/(x)`
`"or "xf(x)=Alog_(e)x+log C" [On integration]"`
Putting x=1, we get
`f(1)=A log_(e)1+log C`
`"or "0=logC" "[becausef(1)=0]`
`"or "xf(x)=A log_(e)x`
Putting x=e, we get
`ef(e)=A log_(e)e`
`"or "A=1" "[becausef(e)=(1)/(e)]`
`therefore" "xf(x)=log_(e)x`
`"or "f(x)(log_(e)x)/(x)`
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