Let f be a differentiable function satisfying the functional time f(xy)=f(x)+f(y)+x+y−1xy∀x,y>0 and f′(1)=2
My work
Putting y=1
f(1)=−1
f′(x)=limh→0f(x+h)−f(x)h
But I don't know anything about f(x+h) so what to do in this problem ?
Answer
Differentiate both sides with respect to x:
yf′(xy)=f′(x)−1x2+1x2y
For y=1/x, we get
f′(1)x=f′(x)−1x2+1x
so
f′(x)=1x2+f′(1)−1x
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