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

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

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

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

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derivative of

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

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

Solution steps

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

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

Simplify

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

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

x \frac{9}{6}, 7

\int_{0}^{2} (1 + x^{2}) d x

\frac{\partial}{\partial x} (\frac{x ^{2}}{16 - y ^{2}})

y^{'}, x^{2} - y^{2} = 9

\frac{d}{d x} (3 cos (2 x) + sin (2 x) + 4)

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