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

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

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

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

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derivative of

y = \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}

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

\frac{x}{2} ln (\frac{1 + x}{1 - x}) - 1

\frac{\partial}{\partial x} ((x - 1) y^{2} e^{x y})

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

n = 1 \sum \infty \frac{n !}{n ^{2} + 1}

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