Question

differentiate please.. y= √x +6 ÷ √x-6

Answer 1

$\bf y=\cfrac{\sqrt{x}+6}{\sqrt{x}-6}\implies \cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{\frac{1}{2}x^{-\frac{1}{2}}(\sqrt{x-6})~~-~~(\sqrt{x}+6)\frac{1}{2}x^{-\frac{1}{2}}}{(\sqrt{x}-6)^2}}$

$\bf \cfrac{dy}{dx}=\cfrac{\frac{1}{2}x^{-\frac{1}{2}}~(\underline{\sqrt{x}}-6~-~\underline{\sqrt{x}}-6)}{(\sqrt{x}-6)^2}\implies \cfrac{dy}{dx}=\cfrac{\frac{1}{2}x^{-\frac{1}{2}}(-12)}{(\sqrt{x}-6)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{\frac{-12}{2\sqrt{x}}}{(\sqrt{x}-6)^2}\implies \cfrac{dy}{dx}=\cfrac{\frac{-6}{\sqrt{x}}}{(\sqrt{x}-6)^2}\implies \cfrac{dy}{dx}=\cfrac{-6}{\sqrt{x}(\sqrt{x}-6)^2}$