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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\frac{d}{d x} (\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}

n = 1 \sum \infty 4 (0.8)^{n - 1}

d x + e^{3 x} d y = 0

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

\frac{d}{d x} (\frac{- 10}{x ^{3}})

e^{x} y d y - (e^{- y} + e^{2 x - y}) d x = 0

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))