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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

l a pl a ce t r an s f or m 0.025 sin (\frac{2 π t}{10})

\frac{d}{d x} (\frac{x}{( x - 2 ) ^{2}})

\int_{1}^{l n (8)} \int_{0}^{l n (y)} e^{x + y} d x d y

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{1}{2}}} d x

a re a x^{4} - 9 x^{2}, x^{2} - 9, 0, 3

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