17. 0 / 2 points\nfind ( y ) and ( y ).\n( y=sqrt{x} ln (x) )\n( y= )\n( y= )
17. 0 / 2 points\nfind ( y ) and ( y ).\n( y=sqrt{x} ln (x) )\n( y= )\n( y= )
Answer
Explanation:
Step1: Find the first derivative using the product rule
The product rule is ((uv)^\prime = u^\prime v+uv^\prime). Let (u = \sqrt{x}=x^{\frac{1}{2}}), then (u^\prime=\frac{1}{2}x^{-\frac{1}{2}}), and (v = \ln(x)), (v^\prime=\frac{1}{x}). [ \begin{align*} y^\prime&=(x^{\frac{1}{2}}\ln(x))^\prime\ &=\frac{1}{2}x^{-\frac{1}{2}}\ln(x)+x^{\frac{1}{2}}\cdot\frac{1}{x}\ &=\frac{\ln(x)}{2\sqrt{x}}+\frac{1}{\sqrt{x}}\ &=\frac{\ln(x) + 2}{2\sqrt{x}} \end{align*} ]
Step2: Find the second derivative using the quotient rule
The quotient rule is ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Let (u=\ln(x)+2), (u^\prime=\frac{1}{x}), and (v = 2\sqrt{x}=2x^{\frac{1}{2}}), (v^\prime=x^{-\frac{1}{2}}). [ \begin{align*} y^{\prime\prime}&=\frac{\frac{1}{x}\cdot2\sqrt{x}-(\ln(x)+2)\cdot x^{-\frac{1}{2}}}{(2\sqrt{x})^{2}}\ &=\frac{\frac{2}{\sqrt{x}}-\frac{\ln(x)+2}{\sqrt{x}}}{4x}\ &=\frac{\frac{2-(\ln(x)+2)}{\sqrt{x}}}{4x}\ &=\frac{-\ln(x)}{4x\sqrt{x}} \end{align*} ]
Answer:
(y^\prime=\frac{\ln(x)+2}{2\sqrt{x}}), (y^{\prime\prime}=\frac{-\ln(x)}{4x\sqrt{x}})