What is the derivative of x^{8/x} ?

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

What is the first derivative of x^{8/x} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to π lim (ln (cos (x) - 8))

cos (5 t)

x \to 2 lim (\frac{x}{( - x + 2 ) ^{2}})

\frac{d y}{d x} = \frac{y ^{2} + 5 x x ^{2} + y ^{2}}{x y}

9 y^{^{''}} - 18 y^{^{'}} + 11 y = 0

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