What is (\partial)/(\partial z)(2x^{y/z}) ?

The answer to (\partial)/(\partial z)(2x^{y/z}) is -(2yx^{y/z}ln(x))/(z^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial z)(2x^{y/z})

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

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

Solution steps

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

Treat

x, y

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

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

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

Apply the chain rule:

e^{\frac{y}{z} l n (x)} \frac{\partial}{\partial z} (\frac{y}{z} ln (x))

= e^{\frac{y}{z} l n (x)} \frac{\partial}{\partial z} (\frac{y}{z} ln (x))

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

= 2 e^{\frac{y}{z} l n (x)} (- \frac{y ln ( x )}{z ^{2}})

Simplify

2 e^{\frac{y}{z} l n (x)} (- \frac{y ln ( x )}{z ^{2}}) : - \frac{2 y x ^{\frac{y}{z}} ln ( x )}{z ^{2}}

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

\int 4 x e^{- x} d x

y = 4 x

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

f (x) = \frac{x}{x ^{2} - 9}

y = x^{2} + 5 x + 6, at x = 7

What is (\partial)/(\partial z)(2x^{y/z}) ?
The answer to (\partial)/(\partial z)(2x^{y/z}) is -(2yx^{y/z}ln(x))/(z^2)