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

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

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

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

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derivative of

y = \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} (ln (x - 4 y))

x \to 1 lim (2 x^{3} + x - 1)

f (x) =, \frac{1}{x}

f (x) = - 3 tan (x) - e^{- x}

y^{^{''}} - 3 y^{^{'}} = (3 t + 8) sin (3 t)

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