What is (\partial)/(\partial x)(x^{-7}y^2+xy^2+2xy) ?

The answer to (\partial)/(\partial x)(x^{-7}y^2+xy^2+2xy) is -(7y^2)/(x^8)+y^2+2y

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(x^{-7}y^2+xy^2+2xy)

\frac{\partial}{\partial x} (x^{- 7} y^{2} + x y^{2} + 2 x y)

- \frac{7 y ^{2}}{x ^{8}} + y^{2} + 2 y

Solution steps

\frac{\partial}{\partial x} (x^{- 7} y^{2} + x y^{2} + 2 x y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (x^{- 7} y^{2}) + \frac{\partial}{\partial x} (x y^{2}) + \frac{\partial}{\partial x} (2 x y)

\frac{\partial}{\partial x} (x^{- 7} y^{2}) = - \frac{7 y ^{2}}{x ^{8}}

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

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

= - \frac{7 y ^{2}}{x ^{8}} + y^{2} + 2 y

n = 0 \sum \infty (2 n)!

x \to - 2 lim (9 x^{2})

x \to - 3 lim (\frac{x ^{2}}{9 - x ^{2}})

y^{^{'}} + \frac{y}{x} = x y^{2}

(2 x - 1) d x (3 y + 7) d y = 0

What is (\partial)/(\partial x)(x^{-7}y^2+xy^2+2xy) ?
The answer to (\partial)/(\partial x)(x^{-7}y^2+xy^2+2xy) is -(7y^2)/(x^8)+y^2+2y