What is (\partial)/(\partial y)((x^2)/(y^2+1)) ?

The answer to (\partial)/(\partial y)((x^2)/(y^2+1)) is -(2x^2y)/((y^2+1)^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^2)/(y^2+1))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial y} (5 x^{2} y)

\frac{d}{d x} (\frac{x ^{2} - 4 x + 20}{x ^{2} + 6 x + 8})

f (x) = \frac{2 x ^{2} + 4 x + 4}{x}

y = \frac{1}{6} x^{3} + \frac{2}{5} x^{2}

\int - sin (e^{x}) e^{2 x} d x

What is (\partial)/(\partial y)((x^2)/(y^2+1)) ?
The answer to (\partial)/(\partial y)((x^2)/(y^2+1)) is -(2x^2y)/((y^2+1)^2)