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

The derivative of (8x^2+6x+6/(sqrt(x))) is (12x^2+3x-3)/(xsqrt(x))

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

The first derivative of (8x^2+6x+6/(sqrt(x))) is (12x^2+3x-3)/(xsqrt(x))

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

\frac{12 x ^{2} + 3 x - 3}{x x}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (8 x^{2} + 6 x + 6) = 16 x + 6

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

= \frac{( 16 x + 6 ) x - \frac{1}{2 x} ( 8 x ^{2} + 6 x + 6 )}{( x ) ^{2}}

Simplify

\frac{( 16 x + 6 ) x - \frac{1}{2 x} ( 8 x ^{2} + 6 x + 6 )}{( x ) ^{2}} : \frac{12 x ^{2} + 3 x - 3}{x x}

= \frac{12 x ^{2} + 3 x - 3}{x x}

\int 16 sin^{2} (x co) s^{2} x d x

y^{^{''}} + 4 y = sec^{2} (2 x)

\int 5 tan^{3} (θ) d θ

\frac{1}{s ^{2} + s + 1}

\int sin^{2} (\frac{1}{2} x) d x

What is the derivative of (8x^2+6x+6/(sqrt(x))) ?
The derivative of (8x^2+6x+6/(sqrt(x))) is (12x^2+3x-3)/(xsqrt(x))
What is the first derivative of (8x^2+6x+6/(sqrt(x))) ?
The first derivative of (8x^2+6x+6/(sqrt(x))) is (12x^2+3x-3)/(xsqrt(x))