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

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

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

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

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derivative of

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

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

Solution steps

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

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

Simplify

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

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

\int_{0}^{3} \frac{1}{3 - x} d x

\int \frac{1}{2 1 - x ^{2}} d x

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 e^{- 2 t}

x - 1, 0

x \to 0 lim (\frac{2}{3 + 4 ^{\frac{1}{0}}})

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