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

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

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

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

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derivative of

\frac{t ^{2} - 6 t}{( t - 3 ) ^{2}}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int sec (t) (3 sec (t) + 5 tan (t)) d t

((x^{2}) 2^{x})^{3}

\frac{d}{d x} (150 - \frac{1}{40} (x - 50)^{2})

\int \frac{1}{x - x + 2} d x

y = x^{2}, y = 4^{\frac{2}{3}}

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