What is the derivative of (8+x^{1/x}) ?

The derivative of (8+x^{1/x}) is (-(ln(8+x))/(x^2)+1/(x(8+x)))(8+x)^{1/x}

What is the first derivative of (8+x^{1/x}) ?

The first derivative of (8+x^{1/x}) is (-(ln(8+x))/(x^2)+1/(x(8+x)))(8+x)^{1/x}

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\frac{d}{d x} ((8 + x)^{\frac{1}{x}})

(- \frac{ln ( 8 + x )}{x ^{2}} + \frac{1}{x ( 8 + x )}) (8 + x)^{\frac{1}{x}}

Solution steps

\frac{d}{d x} ((8 + x)^{\frac{1}{x}})

Apply exponent rule:

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

(8 + x)^{\frac{1}{x}} = e^{\frac{1}{x} l n (8 + x)}

= \frac{d}{d x} (e^{\frac{1}{x} l n (8 + x)})

Apply the chain rule:

e^{\frac{1}{x} l n (8 + x)} \frac{d}{d x} (\frac{1}{x} ln (8 + x))

= e^{\frac{1}{x} l n (8 + x)} \frac{d}{d x} (\frac{1}{x} ln (8 + x))

\frac{d}{d x} (\frac{1}{x} ln (8 + x)) = - \frac{ln ( 8 + x )}{x ^{2}} + \frac{1}{x ( 8 + x )}

= e^{\frac{1}{x} l n (8 + x)} (- \frac{ln ( 8 + x )}{x ^{2}} + \frac{1}{x ( 8 + x )})

e^{\frac{1}{x} l n (8 + x)} = (8 + x)^{\frac{1}{x}}

= (- \frac{ln ( 8 + x )}{x ^{2}} + \frac{1}{x ( 8 + x )}) (8 + x)^{\frac{1}{x}}

\int_{0}^{1} \frac{x ^{7}}{x ^{8} - 1} d x

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

tan (x), \frac{π}{4}

y^{^{'}} = \frac{2 - e ^{x}}{3 + 2 y}, y (0) = 0

y = \frac{1}{x + k e ^{x}}

What is the derivative of (8+x^{1/x}) ?
The derivative of (8+x^{1/x}) is (-(ln(8+x))/(x^2)+1/(x(8+x)))(8+x)^{1/x}
What is the first derivative of (8+x^{1/x}) ?
The first derivative of (8+x^{1/x}) is (-(ln(8+x))/(x^2)+1/(x(8+x)))(8+x)^{1/x}