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

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

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

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

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derivative of

f (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}

y = \frac{1}{6} x^{3} + \frac{2}{5} x^{2}

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

(- 5, 4), (2, 4)

y = 2 x^{2}, y = 0, x = 2

\int e^{- 0.05 x} d x

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