What is the derivative of (6x^2+8x+2)/(sqrt(x)) ?

The derivative of (6x^2+8x+2)/(sqrt(x)) is (9x^2+4x-1)/(xsqrt(x))

What is the first derivative of (6x^2+8x+2)/(sqrt(x)) ?

The first derivative of (6x^2+8x+2)/(sqrt(x)) is (9x^2+4x-1)/(xsqrt(x))

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derivative of

\frac{6 x ^{2} + 8 x + 2}{x}

\frac{9 x ^{2} + 4 x - 1}{x x}

Solution steps

\frac{d}{d x} (\frac{6 x ^{2} + 8 x + 2}{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} + 8 x + 2 ) x - \frac{d}{d x} ( x ) ( 6 x ^{2} + 8 x + 2 )}{( x ) ^{2}}

\frac{d}{d x} (6 x^{2} + 8 x + 2) = 12 x + 8

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

= \frac{( 12 x + 8 ) x - \frac{1}{2 x} ( 6 x ^{2} + 8 x + 2 )}{( x ) ^{2}}

Simplify

\frac{( 12 x + 8 ) x - \frac{1}{2 x} ( 6 x ^{2} + 8 x + 2 )}{( x ) ^{2}} : \frac{9 x ^{2} + 4 x - 1}{x x}

= \frac{9 x ^{2} + 4 x - 1}{x x}

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

x \to 0 lim (+ (\frac{ln ( x )}{x}))

\frac{\partial}{\partial x} (x e^{- x})

\int (9 - y) y d y

n = 1 \sum \infty (cos (24))^{n}

What is the derivative of (6x^2+8x+2)/(sqrt(x)) ?
The derivative of (6x^2+8x+2)/(sqrt(x)) is (9x^2+4x-1)/(xsqrt(x))
What is the first derivative of (6x^2+8x+2)/(sqrt(x)) ?
The first derivative of (6x^2+8x+2)/(sqrt(x)) is (9x^2+4x-1)/(xsqrt(x))