Salut,
[tex]\left(\dfrac{ln(x+1)}{lnx}\right)^{'}=\dfrac{(ln(x+1)^{'}\cdot lnx-ln(x+1)\cdot(lnx)^{'}}{ln^2x}=\\\\\\=\dfrac{\dfrac{1}{x+1}\cdot lnx-\dfrac{1}x\cdot ln(x+1)}{ln^2x}=\dfrac{x\cdot lnx-(x+1)\cdot ln(x+1)}{x\cdot (x+1)\cdot ln^2x}.[/tex]
Ai înțeles rezolvarea ?
Green eyes.
