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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

\int sin (t^{2}) d t

f (x) = 4 x^{2} - 5 x, at x = 3

f (x) = - 4 x

\frac{\partial}{\partial y} (\frac{t}{2 y ^{4}})

\int x sin (5 x^{2}) cos (6 x^{2}) d x

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