What is the derivative of (2x^2+4x+4/(sqrt(x))) ?

The derivative of (2x^2+4x+4/(sqrt(x))) is (3x^2+2x-2)/(xsqrt(x))

What is the first derivative of (2x^2+4x+4/(sqrt(x))) ?

The first derivative of (2x^2+4x+4/(sqrt(x))) is (3x^2+2x-2)/(xsqrt(x))

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

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

Solution steps

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{1 - 4 x - x ^{2}}{x ^{2} - 5})

\int sin^{3} (2 x) cos^{7} (x) d x

\int \frac{1}{x ^{2} - 6 x - 7} d x

x \to 0 lim (\frac{sin ( 5 x )}{x} + 4)

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

What is the derivative of (2x^2+4x+4/(sqrt(x))) ?
The derivative of (2x^2+4x+4/(sqrt(x))) is (3x^2+2x-2)/(xsqrt(x))
What is the first derivative of (2x^2+4x+4/(sqrt(x))) ?
The first derivative of (2x^2+4x+4/(sqrt(x))) is (3x^2+2x-2)/(xsqrt(x))