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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{6 x ^{2} + 4 x - 4}{x x}

\frac{d}{d x} (arctan (\frac{x}{b}))

\int sec^{2} (\frac{x}{2}) d x

f (x) = - 2 x^{3}, at x = - 1

(1, - \frac{1}{2}), (- 2, - \frac{7}{2})

\int_{0}^{\frac{π}{2}} x^{3} cos (2 x) d x

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