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

The derivative of (1+1/x ^{1/x}) is (-(ln(1+1/x))/(x^2)-1/(x^2(x+1)))(1+1/x)^{1/x}

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

The first derivative of (1+1/x ^{1/x}) is (-(ln(1+1/x))/(x^2)-1/(x^2(x+1)))(1+1/x)^{1/x}

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

(- \frac{ln ( 1 + \frac{1}{x} )}{x ^{2}} - \frac{1}{x ^{2} ( x + 1 )}) (1 + \frac{1}{x})^{\frac{1}{x}}

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{1}{x} ln (1 + \frac{1}{x})) = - \frac{ln ( 1 + \frac{1}{x} )}{x ^{2}} - \frac{1}{x ^{2} ( x + 1 )}

= e^{\frac{1}{x} l n (1 + \frac{1}{x})} (- \frac{ln ( 1 + \frac{1}{x} )}{x ^{2}} - \frac{1}{x ^{2} ( x + 1 )})

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

= (- \frac{ln ( 1 + \frac{1}{x} )}{x ^{2}} - \frac{1}{x ^{2} ( x + 1 )}) (1 + \frac{1}{x})^{\frac{1}{x}}

\frac{\partial}{\partial x} (x y x^{2} + y^{2})

\int 3 x sin (5 x) d x

x = 9 y^{2}, x = 28 + 2 y^{2}

\int - e^{t} d t

f (x) = 3 - 6 x^{2}, - 57, at x = 3

What is the derivative of (1+1/x ^{1/x}) ?
The derivative of (1+1/x ^{1/x}) is (-(ln(1+1/x))/(x^2)-1/(x^2(x+1)))(1+1/x)^{1/x}
What is the first derivative of (1+1/x ^{1/x}) ?
The first derivative of (1+1/x ^{1/x}) is (-(ln(1+1/x))/(x^2)-1/(x^2(x+1)))(1+1/x)^{1/x}