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

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

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

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

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derivative of

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

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

Solution steps

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

\frac{d}{d x} (6 x^{2} + 4 x + 4) = 12 x + 4

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

= \frac{( 12 x + 4 ) x - \frac{1}{2 x} ( 6 x ^{2} + 4 x + 4 )}{( x ) ^{2}}

Simplify

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

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

\int_{- 1}^{4} 4 - x^{2} + 3 x d x

s l o p e 6 x^{2} + 3 x y + y^{2} = 16, (1, 2)

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

\int \frac{1}{e ^{2 x} + 4} d x

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{- x}

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