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

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

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

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

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derivative of

y = \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}

\frac{d y}{d x} - 2 x y = 10 x

\frac{\partial}{\partial y} (x^{2} + x y + y^{2} + 2 x - y)

\frac{\partial}{\partial x} (3 x y)

\int cos (4 x) cos (6 x) d x

\int \frac{1}{( x ^{2} + R ^{2} ) ^{\frac{3}{2}}} d x

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