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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x) = e^{x - 1} - 1

y^{^{'}} - 3 y = \frac{y}{x}

\int 6 x^{2} - 4 x d x

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

\frac{\partial}{\partial x} (- 2 e^{x} ln (y))

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