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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x^{2} - 2 x - 3) = 2 x - 2

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

= \frac{( 2 x - 2 ) ( x + 2 ) - 1 \cdot ( x ^{2} - 2 x - 3 )}{( x + 2 ) ^{2}}

Simplify

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

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

\int sin^{2} (v) d v

x \to 0 lim ((\frac{5}{6})^{\frac{1}{6}})

y^{^{'}} - x yy = 2 x y

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

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

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