What is the derivative of (x^3-4)/(x^2+2) ?

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

What is the first derivative of (x^3-4)/(x^2+2) ?

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

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derivative of

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

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

Solution steps

\frac{d}{d x} (\frac{x ^{3} - 4}{x ^{2} + 2})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{3} - 4 ) ( x ^{2} + 2 ) - \frac{d}{d x} ( x ^{2} + 2 ) ( x ^{3} - 4 )}{( x ^{2} + 2 ) ^{2}}

\frac{d}{d x} (x^{3} - 4) = 3 x^{2}

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

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

Expand

3 x^{2} (x^{2} + 2) - 2 x (x^{3} - 4) : x^{4} + 6 x^{2} + 8 x

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

f (t) = \frac{1}{t}

x^{2} + 5 x

x ln (2 x) y^{^{'}} - y = 0

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

(- 4, - 2), (8, 7)

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