determine-the-second-derivative-of-y2x1yleft2sqrtxright-1

Question

Determine the second derivative of y=(2+x)−1y=\left(2+\sqrt{x}\$$right)^{-1}.$$

Accepted Answer

y′$$y^{\prime}$$′=$$2+3x4xx(2+x)3^{\prime}$$=\frac{2+3\sqrt{x}}{4x\sqrt{x}\left(2+\sqrt{x}\$$right)^3$$}

Answer

y′$$y^{\prime}$$′=$$2+3x2xx(2+x)2^{\prime}$$=\frac{2+3\sqrt{x}}{2x\sqrt{x}\left(2+\sqrt{x}\$$right)^2$$}

Answer

y′$$y^{\prime}$$′=−2−$$3x4x(2+x)3^{\prime}$$=-\frac{2-3\sqrt{x}}{4x\left(2+\sqrt{x}\$$right)^3$$}

Answer

y′$$y^{\prime}$$′=−2−$$3x2x(2+x)2^{\prime}$$=-\frac{2-3\sqrt{x}}{2x\left(2+\sqrt{x}\$$right)^2$$}

Determine the second derivative of y=(2+x)−1y=\(\left\)(2+\(\sqrt{x}\]\right\))^{-1}.