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

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

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

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

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derivative of

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

\frac{9 x ^{2} + x - 3}{x x}

Solution steps

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

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

Simplify

\frac{( 12 x + 2 ) x - \frac{1}{2 x} ( 6 x ^{2} + 2 x + 6 )}{( x ) ^{2}} : \frac{9 x ^{2} + x - 3}{x x}

= \frac{9 x ^{2} + x - 3}{x x}

\int r^{2} (r^{3} + 6)^{3} d r

\frac{\partial}{\partial x} (cos^{9} (x^{7} y^{9}))

y = 7 arctan (x + 1 + x^{2})

y = 5 x, y = \frac{5}{2} x, y = 84 - x^{2}

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