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

\frac{\partial}{\partial z} (x y e^{x})

\frac{\partial}{\partial x} ((2 t - 1) (3 t - 6)^{- 1})

\int \frac{x + 1}{x} d x

\frac{d}{d x} (b y (x)^{2})

\int (1 + 2 x)^{4} (2) d 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))