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

The derivative of (2x^2+8x+4/(sqrt(x))) is (3x^2+4x-2)/(xsqrt(x))

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

The first derivative of (2x^2+8x+4/(sqrt(x))) is (3x^2+4x-2)/(xsqrt(x))

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

\frac{3 x ^{2} + 4 x - 2}{x x}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 4 x + 8 ) x - \frac{1}{2 x} ( 2 x ^{2} + 8 x + 4 )}{( x ) ^{2}} : \frac{3 x ^{2} + 4 x - 2}{x x}

= \frac{3 x ^{2} + 4 x - 2}{x x}

y^{^{''}} + 31 y = 0

\frac{d}{d x} (- 8 x^{\frac{1}{2}})

5 x + 6, at x = 6

\int x e^{- s x} d x

\frac{\partial}{\partial x} (8 x^{5} y^{3} - 9 x^{4} y^{6})

What is the derivative of (2x^2+8x+4/(sqrt(x))) ?
The derivative of (2x^2+8x+4/(sqrt(x))) is (3x^2+4x-2)/(xsqrt(x))
What is the first derivative of (2x^2+8x+4/(sqrt(x))) ?
The first derivative of (2x^2+8x+4/(sqrt(x))) is (3x^2+4x-2)/(xsqrt(x))