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

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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

f (x) = 6 x e^{x}

\frac{d}{d x} ((x^{4} - 1) (x^{2} + 1))

\int (8 x^{5} - 8 x^{- 3}) d x

e^{x y}

\frac{10}{s ( s ^{2} + 10 s + 50 )}

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