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

The derivative of ((4x^2+2x+2)/(sqrt(x))) is (6x^2+x-1)/(xsqrt(x))

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

The first derivative of ((4x^2+2x+2)/(sqrt(x))) is (6x^2+x-1)/(xsqrt(x))

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

\frac{6 x ^{2} + x - 1}{x x}

Solution steps

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

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

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

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

Simplify

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

= \frac{6 x ^{2} + x - 1}{x x}

y^{^{''}} + 6 y^{^{'}} = 162 e^{3 t}, y (0) = 6, y^{^{'}} (0) = 7

t \to \infty lim (\frac{0.2 t}{t ^{2} + 1})

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

x \to 5 lim (2 x^{3} - 3 x^{2} - x - 4)

t y^{^{'}} + y = t

What is the derivative of ((4x^2+2x+2)/(sqrt(x))) ?
The derivative of ((4x^2+2x+2)/(sqrt(x))) is (6x^2+x-1)/(xsqrt(x))
What is the first derivative of ((4x^2+2x+2)/(sqrt(x))) ?
The first derivative of ((4x^2+2x+2)/(sqrt(x))) is (6x^2+x-1)/(xsqrt(x))