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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (1.1 x + 6.3 x^{3})

\frac{d y}{d x} + 3 x y^{2} = 0

\int \frac{3 x ^{4} - 2 x ^{3} - 2 x ^{2} + 2 x - 5}{x ^{3} - x ^{2}} d x

y = \frac{2}{x - 4}, (6, 1)

\frac{d y}{d x} = x^{3} (1 - y)

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