What is the derivative of (x^3+1/(x^2-x-2)) ?

The derivative of (x^3+1/(x^2-x-2)) is (x^2-4x+1)/((x-2)^2)

What is the first derivative of (x^3+1/(x^2-x-2)) ?

The first derivative of (x^3+1/(x^2-x-2)) is (x^2-4x+1)/((x-2)^2)

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\frac{d}{d x} (\frac{x ^{3} + 1}{x ^{2} - x - 2})

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

Solution steps

\frac{d}{d x} (\frac{x ^{3} + 1}{x ^{2} - x - 2})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{3} + 1 ) ( x ^{2} - x - 2 ) - \frac{d}{d x} ( x ^{2} - x - 2 ) ( x ^{3} + 1 )}{( x ^{2} - x - 2 ) ^{2}}

\frac{d}{d x} (x^{3} + 1) = 3 x^{2}

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

= \frac{3 x ^{2} ( x ^{2} - x - 2 ) - ( 2 x - 1 ) ( x ^{3} + 1 )}{( x ^{2} - x - 2 ) ^{2}}

Simplify

\frac{3 x ^{2} ( x ^{2} - x - 2 ) - ( 2 x - 1 ) ( x ^{3} + 1 )}{( x ^{2} - x - 2 ) ^{2}} : \frac{x ^{2} - 4 x + 1}{( x - 2 ) ^{2}}

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

\frac{\partial}{\partial y} (tan (2 x + 3 y))

2 - 2 e^{- 4 t}

sin (t) sin (t)

\frac{\partial}{\partial x} (x^{3} - 3 x y + y^{3})

x \to - 2 lim (\frac{1}{∣ x + 2 ∣} + x^{2})

What is the derivative of (x^3+1/(x^2-x-2)) ?
The derivative of (x^3+1/(x^2-x-2)) is (x^2-4x+1)/((x-2)^2)
What is the first derivative of (x^3+1/(x^2-x-2)) ?
The first derivative of (x^3+1/(x^2-x-2)) is (x^2-4x+1)/((x-2)^2)