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

The derivative of ((6x^2+8x+2)/(sqrt(x))) is (9x^2+4x-1)/(xsqrt(x))

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

The first derivative of ((6x^2+8x+2)/(sqrt(x))) is (9x^2+4x-1)/(xsqrt(x))

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

\frac{9 x ^{2} + 4 x - 1}{x x}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (6 x^{2} + 8 x + 2) = 12 x + 8

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

= \frac{( 12 x + 8 ) x - \frac{1}{2 x} ( 6 x ^{2} + 8 x + 2 )}{( x ) ^{2}}

Simplify

\frac{( 12 x + 8 ) x - \frac{1}{2 x} ( 6 x ^{2} + 8 x + 2 )}{( x ) ^{2}} : \frac{9 x ^{2} + 4 x - 1}{x x}

= \frac{9 x ^{2} + 4 x - 1}{x x}

\frac{\partial}{\partial x} (30 y^{4} sin (2 x))

(x - 4) ln (4 - x)

\frac{d}{d x} (x sin (4 x))

f (x) = cos (\frac{π x}{2}), at x = 0.4

\int 30 e^{- x} d x

What is the derivative of ((6x^2+8x+2)/(sqrt(x))) ?
The derivative of ((6x^2+8x+2)/(sqrt(x))) is (9x^2+4x-1)/(xsqrt(x))
What is the first derivative of ((6x^2+8x+2)/(sqrt(x))) ?
The first derivative of ((6x^2+8x+2)/(sqrt(x))) is (9x^2+4x-1)/(xsqrt(x))