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

The derivative of ((4x^2+4x+4)/(sqrt(x))) is (6x^2+2x-2)/(xsqrt(x))

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

The first derivative of ((4x^2+4x+4)/(sqrt(x))) is (6x^2+2x-2)/(xsqrt(x))

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

\frac{6 x ^{2} + 2 x - 2}{x x}

Solution steps

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

\frac{d}{d x} (4 x^{2} + 4 x + 4) = 8 x + 4

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

= \frac{( 8 x + 4 ) x - \frac{1}{2 x} ( 4 x ^{2} + 4 x + 4 )}{( x ) ^{2}}

Simplify

\frac{( 8 x + 4 ) x - \frac{1}{2 x} ( 4 x ^{2} + 4 x + 4 )}{( x ) ^{2}} : \frac{6 x ^{2} + 2 x - 2}{x x}

= \frac{6 x ^{2} + 2 x - 2}{x x}

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

\int_{0}^{\frac{1}{2}} - 2 arccos (x) d x

\int_{0}^{\infty} 9 e^{- 9 x} d x

\frac{\partial}{\partial x} (1 - \frac{3}{y} + x)

\int_{1}^{8} 1 + 3 x d x

What is the derivative of ((4x^2+4x+4)/(sqrt(x))) ?
The derivative of ((4x^2+4x+4)/(sqrt(x))) is (6x^2+2x-2)/(xsqrt(x))
What is the first derivative of ((4x^2+4x+4)/(sqrt(x))) ?
The first derivative of ((4x^2+4x+4)/(sqrt(x))) is (6x^2+2x-2)/(xsqrt(x))