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

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

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

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

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derivative of

f (x) = \frac{3 x ^{2} + 6 x + 4}{x}

\frac{9 x ^{2} + 6 x - 4}{2 x x}

Solution steps

\frac{d}{d x} (\frac{3 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} ( 3 x ^{2} + 6 x + 4 ) x - \frac{d}{d x} ( x ) ( 3 x ^{2} + 6 x + 4 )}{( x ) ^{2}}

\frac{d}{d x} (3 x^{2} + 6 x + 4) = 6 x + 6

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

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

Simplify

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

= \frac{9 x ^{2} + 6 x - 4}{2 x x}

(- 4, \frac{7 π}{6})

8 x + 4 y = 16

y = \frac{x ^{2} + x + 1}{x}

y = x^{4}

y = sin (2 x)

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