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

The derivative of (8x^2+8x+6/(sqrt(x))) is (12x^2+4x-3)/(xsqrt(x))

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

The first derivative of (8x^2+8x+6/(sqrt(x))) is (12x^2+4x-3)/(xsqrt(x))

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

\frac{12 x ^{2} + 4 x - 3}{x x}

Solution steps

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

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

Simplify

\frac{( 16 x + 8 ) x - \frac{1}{2 x} ( 8 x ^{2} + 8 x + 6 )}{( x ) ^{2}} : \frac{12 x ^{2} + 4 x - 3}{x x}

= \frac{12 x ^{2} + 4 x - 3}{x x}

\int \frac{1}{x ^{3} + x ^{2}} d x

\int \frac{29 x ^{4} + 29 x ^{3} + 2 x + 1}{x ^{2} + x} d x

\frac{d}{d x} y = (sin (\frac{x}{2}) - cos (\frac{x}{2}))^{2}

\int \frac{5 - x}{2 x ^{2} + x - 1} d x

\frac{\partial}{\partial x} (a r cot (\frac{y}{x}))

What is the derivative of (8x^2+8x+6/(sqrt(x))) ?
The derivative of (8x^2+8x+6/(sqrt(x))) is (12x^2+4x-3)/(xsqrt(x))
What is the first derivative of (8x^2+8x+6/(sqrt(x))) ?
The first derivative of (8x^2+8x+6/(sqrt(x))) is (12x^2+4x-3)/(xsqrt(x))