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

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

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

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

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derivative of

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

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

Solution steps

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

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

Simplify

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

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

\int \frac{x + 5}{2 x + 3} d x

\int t e^{- 8 t} d t

5 e x cos (x)

\frac{d y}{d x} = e^{x}, x (0) = 6

\frac{\partial}{\partial x} (x^{2} + y^{2} + (12 - x - y)^{2})

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