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

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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

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

f (x) = arcsin (ln (x))

x \to 1 lim (1 - \frac{1}{3 x - 1})

\int \frac{6}{1 - cos ( 2 x )} d x

x y^{^{'}} = y

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