What is the derivative of 2x^{sqrt(x)} ?

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

What is the first derivative of 2x^{sqrt(x)} ?

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{x} = e^{x l n (x)}

= 2 \frac{d}{d x} (e^{x l n (x)})

Apply the chain rule:

e^{x l n (x)} \frac{d}{d x} (x ln (x))

= e^{x l n (x)} \frac{d}{d x} (x ln (x))

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

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

Simplify

2 e^{x l n (x)} (\frac{ln ( x )}{2 x} + \frac{1}{x}) : x^{\frac{2 x - 1}{2}} ln (x) + 2 x^{\frac{2 x - 1}{2}}

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

\int \frac{2 x - 3}{2 x + 1} d x

\int - 4 sin (5 x) sin (x) d x

\frac{d}{d x} (\frac{1}{1 + sin ( x )})

\int \frac{3}{4} t + 6 d t

x \to 3 - lim (9 - x^{2})

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