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

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

x \to 1 lim ((x - 3)^{2})

\frac{\partial}{\partial u} (2 u cos (v))

y^{^{'}} + y - 4 cos (x) = 0

\frac{s}{( s + 2 ) ( s + 3 )}

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

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