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

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

\int \frac{1}{x ^{4}} + 4 d x

\frac{d y}{d x}, x^{3} + y^{3} - 9 x y = 0

f (t) = 1 + e^{4 t}

y = 2 x, (18, 6)

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

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