What is the derivative of (4x^2+6x+2)/(sqrt(x)) ?

The derivative of (4x^2+6x+2)/(sqrt(x)) is (6x^2+3x-1)/(xsqrt(x))

What is the first derivative of (4x^2+6x+2)/(sqrt(x)) ?

The first derivative of (4x^2+6x+2)/(sqrt(x)) is (6x^2+3x-1)/(xsqrt(x))

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derivative of

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

\frac{6 x ^{2} + 3 x - 1}{x x}

Solution steps

\frac{d}{d x} (\frac{4 x ^{2} + 6 x + 2}{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} + 6 x + 2 ) x - \frac{d}{d x} ( x ) ( 4 x ^{2} + 6 x + 2 )}{( x ) ^{2}}

\frac{d}{d x} (4 x^{2} + 6 x + 2) = 8 x + 6

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

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

Simplify

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

= \frac{6 x ^{2} + 3 x - 1}{x x}

\frac{3}{6 t + 8}

cos (x), at x = \frac{π}{4}

\frac{d}{d x} (\frac{g ( x )}{x})

\int - 4 y^{2} d y

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

What is the derivative of (4x^2+6x+2)/(sqrt(x)) ?
The derivative of (4x^2+6x+2)/(sqrt(x)) is (6x^2+3x-1)/(xsqrt(x))
What is the first derivative of (4x^2+6x+2)/(sqrt(x)) ?
The first derivative of (4x^2+6x+2)/(sqrt(x)) is (6x^2+3x-1)/(xsqrt(x))