What is the derivative of ((x^2-6x)/(x+1)) ?

The derivative of ((x^2-6x)/(x+1)) is (x^2+2x-6)/((x+1)^2)

What is the first derivative of ((x^2-6x)/(x+1)) ?

The first derivative of ((x^2-6x)/(x+1)) is (x^2+2x-6)/((x+1)^2)

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\frac{d}{d x} (\frac{( x ^{2} - 6 x )}{x + 1})

\frac{x ^{2} + 2 x - 6}{( x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{( x ^{2} - 6 x )}{x + 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} - 6 x ) ( x + 1 ) - \frac{d}{d x} ( x + 1 ) ( x ^{2} - 6 x )}{( x + 1 ) ^{2}}

\frac{d}{d x} (x^{2} - 6 x) = 2 x - 6

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

= \frac{( 2 x - 6 ) ( x + 1 ) - 1 \cdot ( x ^{2} - 6 x )}{( x + 1 ) ^{2}}

Simplify

\frac{( 2 x - 6 ) ( x + 1 ) - 1 \cdot ( x ^{2} - 6 x )}{( x + 1 ) ^{2}} : \frac{x ^{2} + 2 x - 6}{( x + 1 ) ^{2}}

= \frac{x ^{2} + 2 x - 6}{( x + 1 ) ^{2}}

\int x x^{2} + 9 d x

\int (e^{x} + 2)^{2} d x

\frac{\partial}{\partial x} (- 5 y^{3} sin (5 x))

y = \frac{2}{x}, y = 8 x, y = \frac{1}{8} x

\int_{2}^{\infty} \frac{1}{ln ( x )} d x

What is the derivative of ((x^2-6x)/(x+1)) ?
The derivative of ((x^2-6x)/(x+1)) is (x^2+2x-6)/((x+1)^2)
What is the first derivative of ((x^2-6x)/(x+1)) ?
The first derivative of ((x^2-6x)/(x+1)) is (x^2+2x-6)/((x+1)^2)