What is (\partial)/(\partial y)(x/((x^3+y^2))) ?

The answer to (\partial)/(\partial y)(x/((x^3+y^2))) is -(2xy)/((x^3+y^2)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(x/((x^3+y^2)))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

= x \frac{\partial}{\partial y} (\frac{1}{x ^{3} + y ^{2}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= x \frac{\partial}{\partial y} ((x^{3} + y^{2})^{- 1})

Apply the chain rule:

- \frac{1}{( x ^{3} + y ^{2} ) ^{2}} \frac{\partial}{\partial y} (x^{3} + y^{2})

= - \frac{1}{( x ^{3} + y ^{2} ) ^{2}} \frac{\partial}{\partial y} (x^{3} + y^{2})

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

= x (- \frac{1}{( x ^{3} + y ^{2} ) ^{2}} \cdot 2 y)

Simplify

x (- \frac{1}{( x ^{3} + y ^{2} ) ^{2}} \cdot 2 y) : - \frac{2 x y}{( x ^{3} + y ^{2} ) ^{2}}

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

y = - \frac{e ^{- x}}{2} cos (2 x)

f (x) = x^{3} - 4 x^{2} - 1

\frac{d y}{d t} = - \frac{y}{t} - 8

\frac{d}{d x} (e^{4 x} cos (3 x))

\int sin (4 x) cos (4 x) d x

What is (\partial)/(\partial y)(x/((x^3+y^2))) ?
The answer to (\partial)/(\partial y)(x/((x^3+y^2))) is -(2xy)/((x^3+y^2)^2)