What is the derivative of x^{1/2}ln(x) ?

The derivative of x^{1/2}ln(x) is (ln(x))/(2x^{1/2)}+1/(x^{1/2)}

What is the first derivative of x^{1/2}ln(x) ?

The first derivative of x^{1/2}ln(x) is (ln(x))/(2x^{1/2)}+1/(x^{1/2)}

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\frac{d}{d x} (x^{\frac{1}{2}} ln (x))

\frac{ln ( x )}{2 x ^{\frac{1}{2}}} + \frac{1}{x ^{\frac{1}{2}}}

Solution steps

\frac{d}{d x} (x^{\frac{1}{2}} ln (x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{\frac{1}{2}}, g = ln (x)

= \frac{d}{d x} (x^{\frac{1}{2}}) ln (x) + \frac{d}{d x} (ln (x)) x^{\frac{1}{2}}

\frac{d}{d x} (x^{\frac{1}{2}}) = \frac{1}{2 x ^{\frac{1}{2}}}

\frac{d}{d x} (ln (x)) = \frac{1}{x}

= \frac{1}{2 x ^{\frac{1}{2}}} ln (x) + \frac{1}{x} x^{\frac{1}{2}}

Simplify

\frac{1}{2 x ^{\frac{1}{2}}} ln (x) + \frac{1}{x} x^{\frac{1}{2}} : \frac{ln ( x )}{2 x ^{\frac{1}{2}}} + \frac{1}{x ^{\frac{1}{2}}}

= \frac{ln ( x )}{2 x ^{\frac{1}{2}}} + \frac{1}{x ^{\frac{1}{2}}}

\frac{d}{d x} ((2 x^{2} - 5)^{- 12})

\frac{\partial}{\partial u} (\frac{( u + v )}{2})

\frac{\partial}{\partial x} (x^{2} y cos (\frac{z}{t}))

\frac{\partial}{\partial x} (\frac{- 3 x - y}{y ^{2} + 2 x ^{2}})

\frac{\partial}{\partial x} (e^{2^{x}})

What is the derivative of x^{1/2}ln(x) ?
The derivative of x^{1/2}ln(x) is (ln(x))/(2x^{1/2)}+1/(x^{1/2)}
What is the first derivative of x^{1/2}ln(x) ?
The first derivative of x^{1/2}ln(x) is (ln(x))/(2x^{1/2)}+1/(x^{1/2)}