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

The derivative of (x^2-4x+2/(x+3)) is (x^2+6x-14)/((x+3)^2)

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

The first derivative of (x^2-4x+2/(x+3)) is (x^2+6x-14)/((x+3)^2)

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

\frac{x ^{2} + 6 x - 14}{( x + 3 ) ^{2}}

Solution steps

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

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

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

\frac{d}{d x} (x + 3) = 1

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

Simplify

\frac{( 2 x - 4 ) ( x + 3 ) - 1 \cdot ( x ^{2} - 4 x + 2 )}{( x + 3 ) ^{2}} : \frac{x ^{2} + 6 x - 14}{( x + 3 ) ^{2}}

= \frac{x ^{2} + 6 x - 14}{( x + 3 ) ^{2}}

\int 1 - 1 d x

\int \frac{1}{4 + x ^{2}} d x

\int x^{3} x + 10 d x

x \to 0 lim (tan (x^{2}))

\frac{d y}{d x} = \frac{2}{16 - x ^{2}}

What is the derivative of (x^2-4x+2/(x+3)) ?
The derivative of (x^2-4x+2/(x+3)) is (x^2+6x-14)/((x+3)^2)
What is the first derivative of (x^2-4x+2/(x+3)) ?
The first derivative of (x^2-4x+2/(x+3)) is (x^2+6x-14)/((x+3)^2)