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

The derivative of (2x^2+4x+6)/(sqrt(x)) is (3x^2+2x-3)/(xsqrt(x))

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

The first derivative of (2x^2+4x+6)/(sqrt(x)) is (3x^2+2x-3)/(xsqrt(x))

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derivative of

\frac{2 x ^{2} + 4 x + 6}{x}

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

Solution steps

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

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

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

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

Simplify

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

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

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

- 4 \frac{d y}{d x} + 12 x^{2} = 0

y = 2 x - 3

y = x + 6, y = x^{3}, x = 0

\frac{d}{d x} (a^{- x})

What is the derivative of (2x^2+4x+6)/(sqrt(x)) ?
The derivative of (2x^2+4x+6)/(sqrt(x)) is (3x^2+2x-3)/(xsqrt(x))
What is the first derivative of (2x^2+4x+6)/(sqrt(x)) ?
The first derivative of (2x^2+4x+6)/(sqrt(x)) is (3x^2+2x-3)/(xsqrt(x))