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}

y = tan (x), at x = \frac{π}{4}

\int_{\frac{1}{7}}^{3} 14 x ln (7 x) d x

\frac{\partial}{\partial x} (\frac{5 y}{x ^{2} + 1})

\frac{d}{d x} (\frac{x + 1}{1 - x})

\int_{- 3}^{3} (\frac{27}{2} - \frac{3}{2} x^{2}) d x

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))