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

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

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

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

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

1

Solution steps

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

Simplify

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d x}{d x} = 1

\frac{d}{d x} (4) = 0

= 1 - 0

Simplify

= 1

x \to \infty lim (\frac{3}{2 x})

\frac{d}{d x} (a x + \frac{a}{a x})

\frac{d}{d x} (44 (2 x^{3} - 5)^{\frac{6}{5}})

\frac{d}{d x} (x^{3} os (x))

n = 7 \sum \infty \frac{n ^{2}}{e ^{n}}

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