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

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

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

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

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derivative of

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

\frac{6 x ^{2} + 3 x - 3}{x x}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (4 x^{2} + 6 x + 6) = 8 x + 6

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

= \frac{( 8 x + 6 ) x - \frac{1}{2 x} ( 4 x ^{2} + 6 x + 6 )}{( x ) ^{2}}

Simplify

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

= \frac{6 x ^{2} + 3 x - 3}{x x}

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\frac{d}{d x} (3 cos (- x^{3} + 3 x + 1))

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