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

The derivative of (x^2-6x/((x-3)^2)) is (18)/((x-3)^3)

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

The first derivative of (x^2-6x/((x-3)^2)) is (18)/((x-3)^3)

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

\frac{18}{( x - 3 ) ^{3}}

Solution steps

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

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

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

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

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

Simplify

\frac{( 2 x - 6 ) ( x - 3 ) ^{2} - 2 ( x - 3 ) ( x ^{2} - 6 x )}{( ( x - 3 ) ^{2} ) ^{2}} : \frac{18}{( x - 3 ) ^{3}}

= \frac{18}{( x - 3 ) ^{3}}

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

(12, - 16), (- 9, 8)

\int (tan (x) (6 cos (x) - 6 sec (x))) d x

y^{^{''}} - y^{^{'}} - 6 y = - 4 e^{2} t, y (0) = 0, y^{^{'}} (0) = 0

\int \frac{x ^{2}}{( x - 1 ) ( x ^{2} + 4 x + 5 )} d x

What is the derivative of (x^2-6x/((x-3)^2)) ?
The derivative of (x^2-6x/((x-3)^2)) is (18)/((x-3)^3)
What is the first derivative of (x^2-6x/((x-3)^2)) ?
The first derivative of (x^2-6x/((x-3)^2)) is (18)/((x-3)^3)