What is the derivative of (sqrt(2x^2-2x+1)/x) ?

The derivative of (sqrt(2x^2-2x+1)/x) is (x-1)/(x^2sqrt(2x^2-2x+1))

What is the first derivative of (sqrt(2x^2-2x+1)/x) ?

The first derivative of (sqrt(2x^2-2x+1)/x) is (x-1)/(x^2sqrt(2x^2-2x+1))

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

(0) (4.3)

\int (cos (x) - sin (x))^{2} d x

n = 0 \sum \infty 2^{- 2 - 2 n}

π \cdot c \cdot \frac{cos ( \frac{π x}{L} )}{L}

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

What is the derivative of (sqrt(2x^2-2x+1)/x) ?
The derivative of (sqrt(2x^2-2x+1)/x) is (x-1)/(x^2sqrt(2x^2-2x+1))
What is the first derivative of (sqrt(2x^2-2x+1)/x) ?
The first derivative of (sqrt(2x^2-2x+1)/x) is (x-1)/(x^2sqrt(2x^2-2x+1))