What is the derivative of x^{x+1} ?

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

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

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

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derivative of

x^{x + 1}

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

cos^{2} (x^{2})

e^{- 3} x

\frac{1}{( a ^{2} + x ^{2} )}

\frac{cos ( x )}{2}

\frac{- 2}{x}

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