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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d y}{d x} = e^{5 x + 3 y}

y^{^{''}} + 4 y = 3 tan (2 x)

\int sin (2 x - 5) d x

\int_{0}^{2 π} 7 x^{2} sin (\frac{1}{4} x) d x

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

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