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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} ((6 x - x^{2})^{\frac{1}{2}})

\int t cos (t) sin (t) d t

\frac{d}{d x} (\frac{2}{7 x ^{3}})

\frac{d}{d x} (\frac{1}{a ^{2} - x ^{2}})

x \to 6 lim (5 x + ∣ x - 6 ∣)

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