What is the derivative of (6x^2+8x+4/(sqrt(x))) ?

The derivative of (6x^2+8x+4/(sqrt(x))) is (9x^2+4x-2)/(xsqrt(x))

What is the first derivative of (6x^2+8x+4/(sqrt(x))) ?

The first derivative of (6x^2+8x+4/(sqrt(x))) is (9x^2+4x-2)/(xsqrt(x))

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\frac{d}{d x} (\frac{6 x ^{2} + 8 x + 4}{x})

\frac{9 x ^{2} + 4 x - 2}{x x}

Solution steps

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

\frac{d}{d x} (6 x^{2} + 8 x + 4) = 12 x + 8

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

= \frac{( 12 x + 8 ) x - \frac{1}{2 x} ( 6 x ^{2} + 8 x + 4 )}{( x ) ^{2}}

Simplify

\frac{( 12 x + 8 ) x - \frac{1}{2 x} ( 6 x ^{2} + 8 x + 4 )}{( x ) ^{2}} : \frac{9 x ^{2} + 4 x - 2}{x x}

= \frac{9 x ^{2} + 4 x - 2}{x x}

\int sin (6 x) cos (3 x) d x

\frac{d}{d x} ((x)^{x + 1})

\frac{d x}{d y} = sin (x + y)

(2 x - 1)^{2} \cdot ln (2 x - 1)

(0.4) (6.8)

What is the derivative of (6x^2+8x+4/(sqrt(x))) ?
The derivative of (6x^2+8x+4/(sqrt(x))) is (9x^2+4x-2)/(xsqrt(x))
What is the first derivative of (6x^2+8x+4/(sqrt(x))) ?
The first derivative of (6x^2+8x+4/(sqrt(x))) is (9x^2+4x-2)/(xsqrt(x))