What is the derivative of (2x^2+6x+8)/(sqrt(x)) ?

The derivative of (2x^2+6x+8)/(sqrt(x)) is (3x^2+3x-4)/(xsqrt(x))

What is the first derivative of (2x^2+6x+8)/(sqrt(x)) ?

The first derivative of (2x^2+6x+8)/(sqrt(x)) is (3x^2+3x-4)/(xsqrt(x))

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derivative of

\frac{2 x ^{2} + 6 x + 8}{x}

\frac{3 x ^{2} + 3 x - 4}{x x}

Solution steps

\frac{d}{d x} (\frac{2 x ^{2} + 6 x + 8}{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 + 8 ) x - \frac{d}{d x} ( x ) ( 2 x ^{2} + 6 x + 8 )}{( x ) ^{2}}

\frac{d}{d x} (2 x^{2} + 6 x + 8) = 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 + 8 )}{( x ) ^{2}}

Simplify

\frac{( 4 x + 6 ) x - \frac{1}{2 x} ( 2 x ^{2} + 6 x + 8 )}{( x ) ^{2}} : \frac{3 x ^{2} + 3 x - 4}{x x}

= \frac{3 x ^{2} + 3 x - 4}{x x}

\int \frac{u}{2} d u

\frac{\partial}{\partial x} (x^{2} sin (y) cos (y))

\frac{d}{d v} (u^{2} - v^{2})

\int_{0}^{2} \frac{1}{x + 2} d x

x \to 0 lim (\frac{( cos ( 11 x ) - cos ( 6 x ) )}{x})

What is the derivative of (2x^2+6x+8)/(sqrt(x)) ?
The derivative of (2x^2+6x+8)/(sqrt(x)) is (3x^2+3x-4)/(xsqrt(x))
What is the first derivative of (2x^2+6x+8)/(sqrt(x)) ?
The first derivative of (2x^2+6x+8)/(sqrt(x)) is (3x^2+3x-4)/(xsqrt(x))