What is the derivative of x^x+1 ?

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

What is the first derivative of x^x+1 ?

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{d}{d x} (1) = 0

= x^{x} (ln (x) + 1) + 0

Simplify

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

\frac{\partial}{\partial x} (5 x^{2} + y + sin (a x y) + 3)

\int \frac{3 x ( 8 - x )}{135} d x

y = 2^{x} ln (5)

\int \frac{x}{y} + \frac{y}{x} d y

\int_{0}^{2} x^{3} 1 + 9 x^{4} d x

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