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

The derivative of (2x^2+2x+2/(sqrt(x))) is (3x^2+x-1)/(xsqrt(x))

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

The first derivative of (2x^2+2x+2/(sqrt(x))) is (3x^2+x-1)/(xsqrt(x))

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

\frac{3 x ^{2} + x - 1}{x x}

Solution steps

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

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

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

= \frac{( 4 x + 2 ) x - \frac{1}{2 x} ( 2 x ^{2} + 2 x + 2 )}{( x ) ^{2}}

Simplify

\frac{( 4 x + 2 ) x - \frac{1}{2 x} ( 2 x ^{2} + 2 x + 2 )}{( x ) ^{2}} : \frac{3 x ^{2} + x - 1}{x x}

= \frac{3 x ^{2} + x - 1}{x x}

ln (ln (3 x))

\frac{d}{d x} (\frac{x + 2}{x ^{2} + 2 x + 4})

\frac{d}{d x} (2 e^{3 x} - e^{2 x})

x^{3} y^{^{'''}} - 6 y = 0

\int 7 arctan (3 y) d y

What is the derivative of (2x^2+2x+2/(sqrt(x))) ?
The derivative of (2x^2+2x+2/(sqrt(x))) is (3x^2+x-1)/(xsqrt(x))
What is the first derivative of (2x^2+2x+2/(sqrt(x))) ?
The first derivative of (2x^2+2x+2/(sqrt(x))) is (3x^2+x-1)/(xsqrt(x))