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

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

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

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

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derivative of

y = \frac{4 x ^{2} + 4 x + 8}{x}

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

Solution steps

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

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

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

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

Simplify

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

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

\int_{0}^{8} x^{2} d x

\frac{\partial}{\partial x} (y - x^{2} - 7 z^{2})

f (x) = x^{2} + 8, at x = 2

\int e^{x} + 2 d x

x \to - \infty lim (x^{2} + 6)

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