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

The derivative of y=(8x^2+6x+2)/(sqrt(x)) is (12x^2+3x-1)/(xsqrt(x))

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

The first derivative of y=(8x^2+6x+2)/(sqrt(x)) is (12x^2+3x-1)/(xsqrt(x))

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derivative of

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

\frac{12 x ^{2} + 3 x - 1}{x x}

Solution steps

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

\frac{d}{d x} (8 x^{2} + 6 x + 2) = 16 x + 6

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

= \frac{( 16 x + 6 ) x - \frac{1}{2 x} ( 8 x ^{2} + 6 x + 2 )}{( x ) ^{2}}

Simplify

\frac{( 16 x + 6 ) x - \frac{1}{2 x} ( 8 x ^{2} + 6 x + 2 )}{( x ) ^{2}} : \frac{12 x ^{2} + 3 x - 1}{x x}

= \frac{12 x ^{2} + 3 x - 1}{x x}

\int_{1}^{5} 5 d x

x \to 0 lim (\frac{1}{x + 2})

x \to 0 + lim (x^{3})

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2}}{1 + e ^{x}})

\frac{s}{( s + 2 ) ( s ^{2} + 4 )}

What is the derivative of y=(8x^2+6x+2)/(sqrt(x)) ?
The derivative of y=(8x^2+6x+2)/(sqrt(x)) is (12x^2+3x-1)/(xsqrt(x))
What is the first derivative of y=(8x^2+6x+2)/(sqrt(x)) ?
The first derivative of y=(8x^2+6x+2)/(sqrt(x)) is (12x^2+3x-1)/(xsqrt(x))