What is the derivative of (8x^2+4x+2)/(sqrt(x)) ?

The derivative of (8x^2+4x+2)/(sqrt(x)) is (12x^2+2x-1)/(xsqrt(x))

What is the first derivative of (8x^2+4x+2)/(sqrt(x)) ?

The first derivative of (8x^2+4x+2)/(sqrt(x)) is (12x^2+2x-1)/(xsqrt(x))

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derivative of

\frac{8 x ^{2} + 4 x + 2}{x}

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

Solution steps

\frac{d}{d x} (\frac{8 x ^{2} + 4 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} + 4 x + 2 ) x - \frac{d}{d x} ( x ) ( 8 x ^{2} + 4 x + 2 )}{( x ) ^{2}}

\frac{d}{d x} (8 x^{2} + 4 x + 2) = 16 x + 4

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

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

Simplify

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

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

f (x) = - 3 x x + 1

\int \frac{2 + 2 x}{x - 4 x ( 1 + x )} d x

\int 2 x + \frac{4}{x} d x

x \to 1 lim (- \frac{1}{x})

\int_{1}^{2} (2 x - 4)^{4} d x

What is the derivative of (8x^2+4x+2)/(sqrt(x)) ?
The derivative of (8x^2+4x+2)/(sqrt(x)) is (12x^2+2x-1)/(xsqrt(x))
What is the first derivative of (8x^2+4x+2)/(sqrt(x)) ?
The first derivative of (8x^2+4x+2)/(sqrt(x)) is (12x^2+2x-1)/(xsqrt(x))