find the derivative of $y=sqrt{x}ln x$.
find the derivative of $y=sqrt{x}ln x$.
Answer
Explanation:
Step1: Apply product rule
The product rule is ((uv)^\prime = u^\prime v+uv^\prime). Let (u = \sqrt{x}=x^{\frac{1}{2}}), (v=\ln x). First find (u^\prime) and (v^\prime). (u^\prime=\frac{1}{2}x^{\frac{1}{2}- 1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}), (v^\prime=\frac{1}{x})
Step2: Substitute into product rule
(y^\prime=u^\prime v + uv^\prime=\frac{1}{2\sqrt{x}}\cdot\ln x+\sqrt{x}\cdot\frac{1}{x}) Simplify the second - term (\sqrt{x}\cdot\frac{1}{x}=x^{\frac{1}{2}}\cdot x^{-1}=x^{\frac{1}{2}-1}=x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}) So (y^\prime=\frac{\ln x}{2\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{\ln x + 2}{2\sqrt{x}})
Answer:
(\frac{\ln x + 2}{2\sqrt{x}})