What is the derivative of sqrt(x^2-y^2) ?

The derivative of sqrt(x^2-y^2) is (x-y\frac{dy)/(dx)}{sqrt(x^2-y^2)}

What is the first derivative of sqrt(x^2-y^2) ?

The first derivative of sqrt(x^2-y^2) is (x-y\frac{dy)/(dx)}{sqrt(x^2-y^2)}

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{2}{2 x - 3} d x

\frac{d y}{d x} = 2 y + x^{2} + 5

\frac{d y}{d x} + (\frac{4}{x}) y = 4 x + 9

l a pl a ce t r an s f or m 2 e^{4 t}

t an g e n t x^{4}

What is the derivative of sqrt(x^2-y^2) ?
The derivative of sqrt(x^2-y^2) is (x-y\frac{dy)/(dx)}{sqrt(x^2-y^2)}
What is the first derivative of sqrt(x^2-y^2) ?
The first derivative of sqrt(x^2-y^2) is (x-y\frac{dy)/(dx)}{sqrt(x^2-y^2)}