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

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

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

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

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

\frac{12 x ^{2} + 3 x - 4}{x x}

Solution steps

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

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

Simplify

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

= \frac{12 x ^{2} + 3 x - 4}{x x}

\int_{\frac{2}{4}}^{\frac{1}{2}} \frac{1}{x ^{5} 16 x ^{2} - 1} d x

x \to 0 - lim (x^{2} \cdot sin (\frac{1}{x}))

\int_{- 2}^{3} \frac{33}{3 - x} d x

\int - 3 cos^{2} (x) sin^{2} (x) d x

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