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

The answer to (\partial)/(\partial x)(1/(sqrt(x^2-y^2))) is -x/((x^2-y^2)^{3/2)}

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

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int 9 y^{2} d y

\int_{0}^{5} \frac{1}{25 + t ^{2}} d t

y = sin (x) cos (x)

\int \frac{1}{x} + x d x

y^{^{'}} - 6 y = 0

What is (\partial)/(\partial x)(1/(sqrt(x^2-y^2))) ?
The answer to (\partial)/(\partial x)(1/(sqrt(x^2-y^2))) is -x/((x^2-y^2)^{3/2)}