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

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

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

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

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derivative of

(1 + x)^{x}

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

\frac{3}{cos ( x )}

2 x (1 + 3 x^{2})^{3}

y = (2 x + 1)^{5}

(4 - 3 x^{2})^{5}

g (z) = 10 tan (z) - 2 cot (z)

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