What is (\partial)/(\partial y)(7x^{4y}) ?

The answer to (\partial)/(\partial y)(7x^{4y}) is 28x^{4y}ln(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(7x^{4y})

\frac{\partial}{\partial y} (7 x^{4 y})

28 x^{4 y} ln (x)

Solution steps

\frac{\partial}{\partial y} (7 x^{4 y})

Treat

x

as a constant

Take the constant out:

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

= 7 \frac{\partial}{\partial y} (x^{4 y})

Apply exponent rule:

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

x^{4 y} = e^{4 y l n (x)}

= 7 \frac{\partial}{\partial y} (e^{4 y l n (x)})

Apply the chain rule:

e^{4 y l n (x)} \frac{\partial}{\partial y} (4 y ln (x))

= e^{4 y l n (x)} \frac{\partial}{\partial y} (4 y ln (x))

\frac{\partial}{\partial y} (4 y ln (x)) = 4 ln (x)

= 7 e^{4 y l n (x)} \cdot 4 ln (x)

Simplify

7 e^{4 y l n (x)} \cdot 4 ln (x) : 28 x^{4 y} ln (x)

= 28 x^{4 y} ln (x)

(x - 5) e^{(x)} + 5

f (x) = 6 - x, at x = - 3

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

x^{2} - 4 x - 6, - x^{2} + 4 x + 4

\frac{d y}{d x} = y^{2} \cdot x^{- 3}

What is (\partial)/(\partial y)(7x^{4y}) ?
The answer to (\partial)/(\partial y)(7x^{4y}) is 28x^{4y}ln(x)