What is the derivative of y= t/((t-1)^2) ?

The derivative of y= t/((t-1)^2) is (-1-t)/((t-1)^3)

What is the first derivative of y= t/((t-1)^2) ?

The first derivative of y= t/((t-1)^2) is (-1-t)/((t-1)^3)

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derivative of

y = \frac{t}{( t - 1 ) ^{2}}

\frac{- 1 - t}{( t - 1 ) ^{3}}

Solution steps

\frac{d}{d t} (\frac{t}{( t - 1 ) ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d t}{d t} ( t - 1 ) ^{2} - \frac{d}{d t} ( ( t - 1 ) ^{2} ) t}{( ( t - 1 ) ^{2} ) ^{2}}

\frac{d t}{d t} = 1

\frac{d}{d t} ((t - 1)^{2}) = 2 (t - 1)

= \frac{1 \cdot ( t - 1 ) ^{2} - 2 ( t - 1 ) t}{( ( t - 1 ) ^{2} ) ^{2}}

Simplify

\frac{1 \cdot ( t - 1 ) ^{2} - 2 ( t - 1 ) t}{( ( t - 1 ) ^{2} ) ^{2}} : \frac{- 1 - t}{( t - 1 ) ^{3}}

= \frac{- 1 - t}{( t - 1 ) ^{3}}

f (x) = \frac{4}{5 x + 1}, at x = - 1

(ln (sin (x)))^{^{'}}

\frac{\partial}{\partial x} ((1 x, 3 y, 4 z) \cdot (1, 2, 3))

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

What is the derivative of y= t/((t-1)^2) ?
The derivative of y= t/((t-1)^2) is (-1-t)/((t-1)^3)
What is the first derivative of y= t/((t-1)^2) ?
The first derivative of y= t/((t-1)^2) is (-1-t)/((t-1)^3)