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

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

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

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

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derivative of

f (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{\partial}{\partial y} (x e^{x})

\int (tan^{2} (x) + tan^{4} (x)) d x

n = 2 \sum \infty \frac{2}{3 ^{n}}

\int \frac{2 x}{( x ^{2} + 2 ) ln ( x ^{2} + 2 )} d x

\frac{d y}{d t} - y = 1

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