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

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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

\int_{0}^{1} \frac{30}{6 x ^{2} + 7 x + 1} d x

\int (6 x^{2} - 5 x + 3) d x

2 + 6 e^{a x}

\int_{1}^{2} (- 9 x^{2} + 4 x) d x

x \to 0 lim (\frac{4}{x + 4 h})

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