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

The derivative of ((6x^2+6x+8)/(sqrt(x))) is (9x^2+3x-4)/(xsqrt(x))

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

The first derivative of ((6x^2+6x+8)/(sqrt(x))) is (9x^2+3x-4)/(xsqrt(x))

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

\frac{9 x ^{2} + 3 x - 4}{x x}

Solution steps

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

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

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

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

Simplify

\frac{( 12 x + 6 ) x - \frac{1}{2 x} ( 6 x ^{2} + 6 x + 8 )}{( x ) ^{2}} : \frac{9 x ^{2} + 3 x - 4}{x x}

= \frac{9 x ^{2} + 3 x - 4}{x x}

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4 t \cdot sin (3 t)

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What is the derivative of ((6x^2+6x+8)/(sqrt(x))) ?
The derivative of ((6x^2+6x+8)/(sqrt(x))) is (9x^2+3x-4)/(xsqrt(x))
What is the first derivative of ((6x^2+6x+8)/(sqrt(x))) ?
The first derivative of ((6x^2+6x+8)/(sqrt(x))) is (9x^2+3x-4)/(xsqrt(x))