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

The derivative of (8x^2+6x+4/(sqrt(x))) is (12x^2+3x-2)/(xsqrt(x))

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

The first derivative of (8x^2+6x+4/(sqrt(x))) is (12x^2+3x-2)/(xsqrt(x))

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

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

Solution steps

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

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

Simplify

\frac{( 16 x + 6 ) x - \frac{1}{2 x} ( 8 x ^{2} + 6 x + 4 )}{( x ) ^{2}} : \frac{12 x ^{2} + 3 x - 2}{x x}

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

\frac{d ^{2}}{d x ^{2}} (5 sec (x))

f (x) = 6 e^{x^{- 5}}

2 y d x + 3 x d y = 0

\int e^{2 x} 2 d x

y = x^{2} - 3 x, (5, 10)

What is the derivative of (8x^2+6x+4/(sqrt(x))) ?
The derivative of (8x^2+6x+4/(sqrt(x))) is (12x^2+3x-2)/(xsqrt(x))
What is the first derivative of (8x^2+6x+4/(sqrt(x))) ?
The first derivative of (8x^2+6x+4/(sqrt(x))) is (12x^2+3x-2)/(xsqrt(x))