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

The derivative of f(x)=(2x^2+4x+2)/(sqrt(x)) is (3x^2+2x-1)/(xsqrt(x))

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

The first derivative of f(x)=(2x^2+4x+2)/(sqrt(x)) is (3x^2+2x-1)/(xsqrt(x))

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derivative of

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

\frac{3 x ^{2} + 2 x - 1}{x x}

Solution steps

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

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

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

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

Simplify

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

= \frac{3 x ^{2} + 2 x - 1}{x x}

\frac{\partial}{\partial x} (y + 3 x y^{2} - 18)

f (x) = sin (x), at x = \frac{7 π}{6}

x \to 2 lim (\frac{8 - x ^{3}}{x ^{2} - 2 x})

\int \frac{x + x}{x ^{2}} d x

What is the derivative of f(x)=(2x^2+4x+2)/(sqrt(x)) ?
The derivative of f(x)=(2x^2+4x+2)/(sqrt(x)) is (3x^2+2x-1)/(xsqrt(x))
What is the first derivative of f(x)=(2x^2+4x+2)/(sqrt(x)) ?
The first derivative of f(x)=(2x^2+4x+2)/(sqrt(x)) is (3x^2+2x-1)/(xsqrt(x))