What is the derivative of y=x^{2/x} ?

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

What is the first derivative of y=x^{2/x} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = x^{\frac{2}{x}}

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

Solution steps

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

Apply exponent rule:

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

x^{\frac{2}{x}} = e^{\frac{2}{x} l n (x)}

= \frac{d}{d x} (e^{\frac{2}{x} l n (x)})

Apply the chain rule:

e^{\frac{2}{x} l n (x)} \frac{d}{d x} (\frac{2}{x} ln (x))

= e^{\frac{2}{x} l n (x)} \frac{d}{d x} (\frac{2}{x} ln (x))

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

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

Simplify

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

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

1 t

\frac{\partial}{\partial x} (x^{^{'}} A x)

(- 3, 1), (0, - 3)

\int \frac{- x + 2}{3 ( x ^{2} - x + 1 )} d x

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