What is the derivative of ((2x^2+8x+6)/(sqrt(x))) ?

The derivative of ((2x^2+8x+6)/(sqrt(x))) is (3x^2+4x-3)/(xsqrt(x))

What is the first derivative of ((2x^2+8x+6)/(sqrt(x))) ?

The first derivative of ((2x^2+8x+6)/(sqrt(x))) is (3x^2+4x-3)/(xsqrt(x))

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\frac{d}{d x} (\frac{( 2 x ^{2} + 8 x + 6 )}{x})

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

Solution steps

\frac{d}{d x} (\frac{( 2 x ^{2} + 8 x + 6 )}{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 + 6 ) x - \frac{d}{d x} ( x ) ( 2 x ^{2} + 8 x + 6 )}{( x ) ^{2}}

\frac{d}{d x} (2 x^{2} + 8 x + 6) = 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 + 6 )}{( x ) ^{2}}

Simplify

\frac{( 4 x + 8 ) x - \frac{1}{2 x} ( 2 x ^{2} + 8 x + 6 )}{( x ) ^{2}} : \frac{3 x ^{2} + 4 x - 3}{x x}

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

cos (\frac{1 - e ^{3 x}}{1 + e ^{3 x}})

\frac{\partial}{\partial x} (2 x y^{3} - 4 x^{2} y^{2})

x^{\frac{1}{2}} + y^{\frac{1}{2}} = 4

f (x) = arcsin (4 x)

f (x) = - 6 x^{2} - 10 x, at x = 1

What is the derivative of ((2x^2+8x+6)/(sqrt(x))) ?
The derivative of ((2x^2+8x+6)/(sqrt(x))) is (3x^2+4x-3)/(xsqrt(x))
What is the first derivative of ((2x^2+8x+6)/(sqrt(x))) ?
The first derivative of ((2x^2+8x+6)/(sqrt(x))) is (3x^2+4x-3)/(xsqrt(x))