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

The derivative of y=(8x^2+2x+8)/(sqrt(x)) is (12x^2+x-4)/(xsqrt(x))

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

The first derivative of y=(8x^2+2x+8)/(sqrt(x)) is (12x^2+x-4)/(xsqrt(x))

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derivative of

y = \frac{8 x ^{2} + 2 x + 8}{x}

\frac{12 x ^{2} + x - 4}{x x}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (8 x^{2} + 2 x + 8) = 16 x + 2

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

= \frac{( 16 x + 2 ) x - \frac{1}{2 x} ( 8 x ^{2} + 2 x + 8 )}{( x ) ^{2}}

Simplify

\frac{( 16 x + 2 ) x - \frac{1}{2 x} ( 8 x ^{2} + 2 x + 8 )}{( x ) ^{2}} : \frac{12 x ^{2} + x - 4}{x x}

= \frac{12 x ^{2} + x - 4}{x x}

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

\frac{d}{d x} (\frac{e ^{2 x} - e ^{x} + e ^{- 3 x}}{e ^{x}})

\frac{\partial}{\partial x} (2 x)

\int \frac{1}{x ^{2} + 1} d x

\frac{\partial}{\partial x} (z sin (x))

What is the derivative of y=(8x^2+2x+8)/(sqrt(x)) ?
The derivative of y=(8x^2+2x+8)/(sqrt(x)) is (12x^2+x-4)/(xsqrt(x))
What is the first derivative of y=(8x^2+2x+8)/(sqrt(x)) ?
The first derivative of y=(8x^2+2x+8)/(sqrt(x)) is (12x^2+x-4)/(xsqrt(x))