What is the derivative of ((6x^2+6x+4))/(sqrt(x)) ?

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

What is the first derivative of ((6x^2+6x+4))/(sqrt(x)) ?

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

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derivative of

\frac{( 6 x ^{2} + 6 x + 4 )}{x}

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

Solution steps

\frac{d}{d x} (\frac{6 x ^{2} + 6 x + 4}{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 + 4 ) x - \frac{d}{d x} ( x ) ( 6 x ^{2} + 6 x + 4 )}{( x ) ^{2}}

\frac{d}{d x} (6 x^{2} + 6 x + 4) = 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 + 4 )}{( x ) ^{2}}

Simplify

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

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

\frac{d}{d x} (x y^{4})

\frac{d}{d x} ((x^{2} - 9)^{3})

\int x^{37} sin (x^{38}) d x

\int sec (\frac{x}{8}) d x

\int_{8}^{\infty} x e^{- 5 x} d x

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