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

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

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

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

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derivative of

f (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}}

4 sin (t)

\frac{d}{d x} (e^{3 x + 5})

\int 2 r (sin (θ)) z^{3}

x^{2}, x^{3}, 0, \frac{1}{4}

n = 1 \sum \infty \frac{1}{n 9 ^{n}}

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