What is the derivative of kx^x ?

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

What is the first derivative of kx^x ?

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

= k \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

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

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

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

\frac{d}{d x} (\frac{x ^{2}}{( 4 + x )})

\frac{d}{d x} (\frac{3 x + 4}{3 x - 4})

y^{2} = x^{2} (x + 1), (3, 6)

\int x^{l n (x)} d x

f (x) = x^{2} - 3 x, at x = - 2

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