What is the derivative of (x^2+6x+9/(x-3)) ?

The derivative of (x^2+6x+9/(x-3)) is (x^2-6x-27)/((x-3)^2)

What is the first derivative of (x^2+6x+9/(x-3)) ?

The first derivative of (x^2+6x+9/(x-3)) is (x^2-6x-27)/((x-3)^2)

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\frac{d}{d x} (\frac{x ^{2} + 6 x + 9}{x - 3})

\frac{x ^{2} - 6 x - 27}{( x - 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 6 x + 9}{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} + 6 x + 9 ) ( x - 3 ) - \frac{d}{d x} ( x - 3 ) ( x ^{2} + 6 x + 9 )}{( x - 3 ) ^{2}}

\frac{d}{d x} (x^{2} + 6 x + 9) = 2 x + 6

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

= \frac{( 2 x + 6 ) ( x - 3 ) - 1 \cdot ( x ^{2} + 6 x + 9 )}{( x - 3 ) ^{2}}

Simplify

\frac{( 2 x + 6 ) ( x - 3 ) - 1 \cdot ( x ^{2} + 6 x + 9 )}{( x - 3 ) ^{2}} : \frac{x ^{2} - 6 x - 27}{( x - 3 ) ^{2}}

= \frac{x ^{2} - 6 x - 27}{( x - 3 ) ^{2}}

f (x) = (2 x - 5) (3 x + 1)

y^{^{'}} = \frac{x y ^{3}}{1 + x ^{2}}, y (0) = 1

\frac{\partial}{\partial x} (\frac{( 3 x - 4 y )}{- x + 6 y})

- \frac{x}{9 y}

(8, 1), (0, - 3)

What is the derivative of (x^2+6x+9/(x-3)) ?
The derivative of (x^2+6x+9/(x-3)) is (x^2-6x-27)/((x-3)^2)
What is the first derivative of (x^2+6x+9/(x-3)) ?
The first derivative of (x^2+6x+9/(x-3)) is (x^2-6x-27)/((x-3)^2)