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}

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = - 3, y^{^{'}} (0) = 1

x \to 3 lim (cos (π x) + 6)

\int \frac{4 3 x}{3} d x

\frac{\partial}{\partial x} (110 e^{- 0.1 t})

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