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

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

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

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

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\frac{d}{d x} (\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 \infty lim (\frac{n !}{n})

\int \frac{x + 1}{x ^{2} + 4 x + 3} d x

\int_{6}^{8} \frac{30}{( x - 6 ) ^{3}} d x

\int_{0}^{x} (x + 2 y) d y

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

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