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

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

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

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

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derivative of

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{e ^{9 x}}{e ^{9 x} + 1} d x

f (x) = sin (x + 1)

\int 3 y^{2} \frac{1}{2 y} tan (2 y x) d x

\int 6 x^{- 1} d x

\int e^{2 + x} d x

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