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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

y = 2 x^{2}, y = 16 x - 32, y = 0

\frac{d y}{d t} = \frac{y + 1}{t + 1}

\int \frac{1}{1 - x ^{2}} d x

\frac{\partial}{\partial x} (6 - arctan (7 x^{2} + 2 y^{2}))

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

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