What is the derivative of (2x^2+6x+6)/(sqrt(x)) ?

The derivative of (2x^2+6x+6)/(sqrt(x)) is (3x^2+3x-3)/(xsqrt(x))

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

The first derivative of (2x^2+6x+6)/(sqrt(x)) is (3x^2+3x-3)/(xsqrt(x))

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derivative of

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

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

Solution steps

\frac{d}{d x} (\frac{2 x ^{2} + 6 x + 6}{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} + 6 x + 6 ) x - \frac{d}{d x} ( x ) ( 2 x ^{2} + 6 x + 6 )}{( x ) ^{2}}

\frac{d}{d x} (2 x^{2} + 6 x + 6) = 4 x + 6

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

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

Simplify

\frac{( 4 x + 6 ) x - \frac{1}{2 x} ( 2 x ^{2} + 6 x + 6 )}{( x ) ^{2}} : \frac{3 x ^{2} + 3 x - 3}{x x}

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

\int x 8 - x^{2} d x

s l o p e in t erce pt (- 1, 19), (5, - 41)

d er i v a t i v e 50 x - x x

\int \frac{2}{3 x x} d x

\int_{0}^{\frac{π}{4}} 2 tan^{4} (x) d x

What is the derivative of (2x^2+6x+6)/(sqrt(x)) ?
The derivative of (2x^2+6x+6)/(sqrt(x)) is (3x^2+3x-3)/(xsqrt(x))
What is the first derivative of (2x^2+6x+6)/(sqrt(x)) ?
The first derivative of (2x^2+6x+6)/(sqrt(x)) is (3x^2+3x-3)/(xsqrt(x))