What is the derivative of 3x^x ?

The derivative of 3x^x is 3x^x(ln(x)+1)

What is the first derivative of 3x^x ?

The first derivative of 3x^x is 3x^x(ln(x)+1)

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\frac{d}{d x} (3 x^{x})

3 x^{x} (ln (x) + 1)

Solution steps

\frac{d}{d x} (3 x^{x})

Take the constant out:

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

= 3 \frac{d}{d x} (x^{x})

Apply exponent rule:

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

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

= 3 \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)) = ln (x) + 1

= 3 e^{x l n (x)} (ln (x) + 1)

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

= 3 x^{x} (ln (x) + 1)

x y + y^{^{'}} = 92 x

\frac{d}{d x} (sin (\frac{x}{2}) - e^{- x^{3}})

\frac{\partial}{\partial x} ((\frac{y}{x - y})^{6})

x^{4^{x}}

What is the derivative of 3x^x ?
The derivative of 3x^x is 3x^x(ln(x)+1)
What is the first derivative of 3x^x ?
The first derivative of 3x^x is 3x^x(ln(x)+1)