What is the first d/(dy)(x^2y^4) ?

The first d/(dy)(x^2y^4) is 4x^2y^3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

d/(dy)(x^2y^4)

\frac{d}{d y} (x^{2} y^{4})

4 x^{2} y^{3}

Solution steps

\frac{d}{d y} (x^{2} y^{4})

Treat

x

as a constant

Take the constant out:

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

= x^{2} \frac{d}{d y} (y^{4})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= x^{2} \cdot 4 y^{4 - 1}

Simplify

= 4 x^{2} y^{3}

n = 0 \sum \infty (\frac{1}{1.01})^{n}

\frac{d y}{d t} t + 5 y = 8 t

\frac{d}{d t} (\frac{t ^{2}}{t ^{3} + 1})

- x^{2} sin (x) + 4 x cos (x) + 2 sin (x)

\int - 8 ln (x^{2} - 1) d x