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

The derivative of (2x^2+6x+2)/(sqrt(x)) is (3x^2+3x-1)/(xsqrt(x))

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

The first derivative of (2x^2+6x+2)/(sqrt(x)) is (3x^2+3x-1)/(xsqrt(x))

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derivative of

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

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

Solution steps

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

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

Simplify

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

= \frac{3 x ^{2} + 3 x - 1}{x x}

\int sec^{2} (x) tan (x) d x

(\frac{1}{x} + 2 y^{2} x) d x + (2 y x^{2} - cos (y)) d y = 0, y (1) = π

n = 1 \sum \infty (sin (3))^{n}

\frac{d}{d x} (e^{x + y} - 1)

\int \frac{sin ^{4} ( x )}{cos ^{6} ( x )} d x

What is the derivative of (2x^2+6x+2)/(sqrt(x)) ?
The derivative of (2x^2+6x+2)/(sqrt(x)) is (3x^2+3x-1)/(xsqrt(x))
What is the first derivative of (2x^2+6x+2)/(sqrt(x)) ?
The first derivative of (2x^2+6x+2)/(sqrt(x)) is (3x^2+3x-1)/(xsqrt(x))