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

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

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

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

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

\frac{12 x ^{2} + 4 x - 1}{x x}

Solution steps

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

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

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

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

Simplify

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

= \frac{12 x ^{2} + 4 x - 1}{x x}

\int \frac{4}{x ^{2} 16 - x ^{2}} d x

(x + 2)^{2} + (y - 3)^{2} = 37, (4, 4)

\int \frac{x}{1} d x

x = 5 y^{2}, x = 1 + 4 y^{2}

3 x^{- \frac{3}{2}} + 6 x^{- \frac{1}{2}} + x^{3} - 4

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