What is the d/(d{z)}(9{x}^{({y})/({z)}}) ?

The d/(d{z)}(9{x}^{({y})/({z)}}) is -(9e^{\frac{ln({x}){y})/({z)}}ln({x}){y}}{{z}^2}

What is the first d/(d{z)}(9{x}^{({y})/({z)}}) ?

The first d/(d{z)}(9{x}^{({y})/({z)}}) is -(9e^{\frac{ln({x}){y})/({z)}}ln({x}){y}}{{z}^2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d z} (9 x^{\frac{y}{z}})

- \frac{9 e ^{\frac{l n ( x ) y}{z}} ln ( x ) y}{z ^{2}}

Solution steps

\frac{d}{d z} (9 x^{\frac{y}{z}})

Take the constant out:

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

= 9 \frac{d}{d 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)}

= 9 \frac{d}{d z} (e^{\frac{y}{z} l n (x)})

Apply the chain rule:

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

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

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

= 9 e^{\frac{y}{z} l n (x)} (- \frac{ln ( x ) y}{z ^{2}})

Simplify

9 e^{\frac{y}{z} l n (x)} (- \frac{ln ( x ) y}{z ^{2}}) : - \frac{9 e ^{\frac{l n ( x ) y}{z}} ln ( x ) y}{z ^{2}}

= - \frac{9 e ^{\frac{l n ( x ) y}{z}} ln ( x ) y}{z ^{2}}

\int (2 x + 1) e^{x^{2} + 1} d x

\frac{1}{( x - 2 )}

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} - 40 y = 1 - 4 t

\int_{2}^{\infty} x e^{- 8 x} d x

e^{x}, x e^{x}, 0, 1

What is the d/(d{z)}(9{x}^{({y})/({z)}}) ?
The d/(d{z)}(9{x}^{({y})/({z)}}) is -(9e^{\frac{ln({x}){y})/({z)}}ln({x}){y}}{{z}^2}
What is the first d/(d{z)}(9{x}^{({y})/({z)}}) ?
The first d/(d{z)}(9{x}^{({y})/({z)}}) is -(9e^{\frac{ln({x}){y})/({z)}}ln({x}){y}}{{z}^2}