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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

1 + \frac{1}{s ( s + 2 )}

\frac{1 - 4 x}{2 + x}

\int_{0}^{1} π (2 x)^{2} d x

\int \frac{2 s - 8}{( s ^{2} - 8 s + 25 ) ^{2}} d s

\frac{9 y ^{2} - t ^{2}}{y ^{5}} \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

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