1=جد المشتقة الثانية اذا علمت ان جذر اكس - جذر واي
I asked my Algerian friend, Majid, to translate this for me.
I think this is correct but I would like another mathematician to check it please.
Find the second derivative knowing that square root(x) - squre root(y) = 1
\(\sqrt{x}-\sqrt{y}=1\\ x^{0.5}-y^{0.5}=1\\ 0.5x^{-0.5}-0.5y^{-0.5}\frac{dy}{dx}=0\\ 0.5x^{-0.5}=0.5y^{-0.5}\frac{dy}{dx}\\ \frac{0.5x^{-0.5}}{0.5y^{-0.5}}=\frac{dy}{dx}\\ \frac{dy}{dx}=\frac{x^{-0.5}}{y^{-0.5}}\\ \frac{dy}{dx}=\frac{y^{0.5}}{x^{0.5}}\\ \frac{d^2y}{dx^2}=\frac{x^{0.5}*0.5y^{-0.5}\frac{dy}{dx}\;-\;y^{0.5}*0.5x^{-0.5}}{x}\\ \dfrac{d^2y}{dx^2}=\dfrac{\frac{x^{0.5}}{2y^{0.5}}\frac{dy}{dx}- \frac{y^{0.5}}{2x^{0.5}} } {x}\\ \dfrac{d^2y}{dx^2}=\dfrac{\frac{x^{0.5}}{2y^{0.5}}\frac{y^{0.5}}{x^{0.5}}- \frac{y^{0.5}}{2x^{0.5}} } {x}\\ \dfrac{d^2y}{dx^2}=\dfrac{\frac{x^{0.5}}{2x^{0.5}}- \frac{y^{0.5}}{2x^{0.5}} } {x}\\ \)
\(\dfrac{d^2y}{dx^2}=\dfrac{x^{0.5}-y^{0.5}}{2x^{0.5}} \div x\\ \dfrac{d^2y}{dx^2}=\dfrac{x^{0.5}-y^{0.5}}{2x^{0.5}*x} \\ \dfrac{d^2y}{dx^2}=\dfrac{x^{0.5}-y^{0.5}}{2x^{1.5}} \\ \dfrac{d^2y}{dx^2}=\dfrac{\sqrt{x}-\sqrt{y}}{2x\sqrt{x}} \\ \)
