What is the derivative of x^{5/x} ?

The derivative of x^{5/x} is 5x^{(5-2x)/x}(-ln(x)+1)

What is the first derivative of x^{5/x} ?

The first derivative of x^{5/x} is 5x^{(5-2x)/x}(-ln(x)+1)

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\frac{d}{d x} (x^{\frac{5}{x}})

5 x^{\frac{5 - 2 x}{x}} (- ln (x) + 1)

Solution steps

\frac{d}{d x} (x^{\frac{5}{x}})

Apply exponent rule:

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

x^{\frac{5}{x}} = e^{\frac{5}{x} l n (x)}

= \frac{d}{d x} (e^{\frac{5}{x} l n (x)})

Apply the chain rule:

e^{\frac{5}{x} l n (x)} \frac{d}{d x} (\frac{5}{x} ln (x))

= e^{\frac{5}{x} l n (x)} \frac{d}{d x} (\frac{5}{x} ln (x))

\frac{d}{d x} (\frac{5}{x} ln (x)) = \frac{5 ( - ln ( x ) + 1 )}{x ^{2}}

= e^{\frac{5}{x} l n (x)} \frac{5 ( - ln ( x ) + 1 )}{x ^{2}}

Simplify

e^{\frac{5}{x} l n (x)} \frac{5 ( - ln ( x ) + 1 )}{x ^{2}} : 5 x^{\frac{5 - 2 x}{x}} (- ln (x) + 1)

= 5 x^{\frac{5 - 2 x}{x}} (- ln (x) + 1)

\int_{0}^{9} x e^{- x^{2}} d x

\frac{1}{x ^{2}}, 1

\frac{\partial}{\partial x} (\frac{x ^{2}}{y} - x z)

\frac{\partial}{\partial y} (2 ln (3 x) - 4 ln (2 x y))

\frac{3}{x ^{3}}

What is the derivative of x^{5/x} ?
The derivative of x^{5/x} is 5x^{(5-2x)/x}(-ln(x)+1)
What is the first derivative of x^{5/x} ?
The first derivative of x^{5/x} is 5x^{(5-2x)/x}(-ln(x)+1)