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

The derivative of (6x^2+4x+6/(sqrt(x))) is (9x^2+2x-3)/(xsqrt(x))

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

The first derivative of (6x^2+4x+6/(sqrt(x))) is (9x^2+2x-3)/(xsqrt(x))

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

\frac{9 x ^{2} + 2 x - 3}{x x}

Solution steps

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

\frac{d}{d x} (6 x^{2} + 4 x + 6) = 12 x + 4

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

= \frac{( 12 x + 4 ) x - \frac{1}{2 x} ( 6 x ^{2} + 4 x + 6 )}{( x ) ^{2}}

Simplify

\frac{( 12 x + 4 ) x - \frac{1}{2 x} ( 6 x ^{2} + 4 x + 6 )}{( x ) ^{2}} : \frac{9 x ^{2} + 2 x - 3}{x x}

= \frac{9 x ^{2} + 2 x - 3}{x x}

\frac{d}{d x} (\frac{x ^{6}}{3 - x ^{5}})

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

x^{^{'}} + 2 x = 5 e^{3 t}

x = y^{2}, x = y + 2, y = 0

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

What is the derivative of (6x^2+4x+6/(sqrt(x))) ?
The derivative of (6x^2+4x+6/(sqrt(x))) is (9x^2+2x-3)/(xsqrt(x))
What is the first derivative of (6x^2+4x+6/(sqrt(x))) ?
The first derivative of (6x^2+4x+6/(sqrt(x))) is (9x^2+2x-3)/(xsqrt(x))