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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

\frac{\partial}{\partial x} (x^{9} y^{6} - x^{8} y^{5})

\frac{d}{d x} (e^{x} + 18)

\int - 16 tan (x) cos (x) d x

x \to 0 lim (\frac{\frac{1}{1 - x} + 1}{x})

x^{^{''}} + \frac{9}{4} x^{^{'}} + \frac{9}{64} x = 0, x^{^{'}} (0) = 1, x (0) = 2

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