What is the derivative of y=x^x ?

The derivative of y=x^x is x^x(ln(x)+1)

What is the first derivative of y=x^x ?

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

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derivative of

y = x^{x}

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

Solution steps

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

Apply exponent rule:

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

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

= \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

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

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

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

y = ln (2 x)

(2, \frac{3 π}{2})

f (x) = 5 x e^{x^{2} - 3}

(5, 53)

(0, 3), (2, - 3)

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