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

The derivative of (4x^2-x/(x^2+2)) is (x^2+16x-2)/((x^2+2)^2)

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

The first derivative of (4x^2-x/(x^2+2)) is (x^2+16x-2)/((x^2+2)^2)

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

\frac{x ^{2} + 16 x - 2}{( x ^{2} + 2 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (4 x^{2} - x) = 8 x - 1

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

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

Expand

(8 x - 1) (x^{2} + 2) - 2 x (4 x^{2} - x) : x^{2} + 16 x - 2

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

\frac{6 x ^{2} + 2 x + 2}{x}

x \to 0 - lim (\frac{x - 1}{x})

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

x \to 0 lim (\frac{cos ( x ^{2} )}{x ^{2}})

y^{^{'}} = 6 + t - y

What is the derivative of (4x^2-x/(x^2+2)) ?
The derivative of (4x^2-x/(x^2+2)) is (x^2+16x-2)/((x^2+2)^2)
What is the first derivative of (4x^2-x/(x^2+2)) ?
The first derivative of (4x^2-x/(x^2+2)) is (x^2+16x-2)/((x^2+2)^2)