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

The derivative of (sqrt(x^2-2x)/(x^2)) is (-x+3)/(x^2sqrt(x^2-2x))

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

The first derivative of (sqrt(x^2-2x)/(x^2)) is (-x+3)/(x^2sqrt(x^2-2x))

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Popular >

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

\frac{- x + 3}{x ^{2} x ^{2} - 2 x}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{- x + 3}{x ^{2} x ^{2} - 2 x}

f (x) = \frac{x}{x ^{2} + 4}

t^{3} y^{^{'}} + 4 t^{2} y = e^{- t}, y (- 1) = 0

\int - 3 sin (x) d x

x (x^{2} + 1)^{10}

\int_{1}^{5} (x - 3) d x

What is the derivative of (sqrt(x^2-2x)/(x^2)) ?
The derivative of (sqrt(x^2-2x)/(x^2)) is (-x+3)/(x^2sqrt(x^2-2x))
What is the first derivative of (sqrt(x^2-2x)/(x^2)) ?
The first derivative of (sqrt(x^2-2x)/(x^2)) is (-x+3)/(x^2sqrt(x^2-2x))