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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\frac{d y}{d x} - 5 y = - \frac{5}{2} x y^{3}

\int \frac{x}{x ^{2} + 6} d x

y^{^{''}} - 9 y^{^{'}} + 18 y = 0, y (0) = - 5, y^{^{'}} (0) = 2

f (x) = 9 x^{2} - x^{3}, (2, 28)

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