What is the derivative of (sqrt(x^2+y^2-9)/x) ?

The derivative of (sqrt(x^2+y^2-9)/x) is (xy\frac{dy)/(dx)-y^2+9}{x^2sqrt(x^2+y^2-9)}

What is the first derivative of (sqrt(x^2+y^2-9)/x) ?

The first derivative of (sqrt(x^2+y^2-9)/x) is (xy\frac{dy)/(dx)-y^2+9}{x^2sqrt(x^2+y^2-9)}

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\frac{d}{d x} (\frac{x ^{2} + y ^{2} - 9}{x})

\frac{x y \frac{d y}{d x} - y ^{2} + 9}{x ^{2} x ^{2} + y ^{2} - 9}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + y ^{2} - 9}{x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( x ^{2} + y ^{2} - 9 ) x - \frac{d x}{d x} x ^{2} + y ^{2} - 9}{x ^{2}}

\frac{d}{d x} (x^{2} + y^{2} - 9) = \frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 9}

\frac{d x}{d x} = 1

= \frac{\frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 9} x - 1 \cdot x ^{2} + y ^{2} - 9}{x ^{2}}

Simplify

\frac{\frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 9} x - 1 \cdot x ^{2} + y ^{2} - 9}{x ^{2}} : \frac{x y \frac{d y}{d x} - y ^{2} + 9}{x ^{2} x ^{2} + y ^{2} - 9}

= \frac{x y \frac{d y}{d x} - y ^{2} + 9}{x ^{2} x ^{2} + y ^{2} - 9}

n = 1 \sum \infty \frac{- 1 ^{n + 1}}{n}

\int x^{2} cos (\frac{1}{2} x) d x

\int_{0}^{1} \frac{6}{1 + t ^{2}} d t

\frac{d}{d x} (\frac{cos ( x )}{x} + \frac{x}{cos ( x )})

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

What is the derivative of (sqrt(x^2+y^2-9)/x) ?
The derivative of (sqrt(x^2+y^2-9)/x) is (xy\frac{dy)/(dx)-y^2+9}{x^2sqrt(x^2+y^2-9)}
What is the first derivative of (sqrt(x^2+y^2-9)/x) ?
The first derivative of (sqrt(x^2+y^2-9)/x) is (xy\frac{dy)/(dx)-y^2+9}{x^2sqrt(x^2+y^2-9)}