What is the d/(dt)(((t+1)^2)/((t-1)^2)) ?

The d/(dt)(((t+1)^2)/((t-1)^2)) is -(4(t+1))/((t-1)^3)

What is the first d/(dt)(((t+1)^2)/((t-1)^2)) ?

The first d/(dt)(((t+1)^2)/((t-1)^2)) is -(4(t+1))/((t-1)^3)

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

- \frac{4 ( t + 1 )}{( t - 1 ) ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{2 ( t + 1 ) ( t - 1 ) ^{2} - 2 ( t - 1 ) ( t + 1 ) ^{2}}{( ( t - 1 ) ^{2} ) ^{2}} : - \frac{4 ( t + 1 )}{( t - 1 ) ^{3}}

= - \frac{4 ( t + 1 )}{( t - 1 ) ^{3}}

\int sin (wt - b x) d t

\frac{\partial}{\partial y} (2 x^{2} + 6 x y + y^{2})

c cos (t) + t^{2} sin (t)

\int (2 + 2 cos (θ))^{2} d θ

\int \frac{1}{4 x ^{2} + 4 x + 10} d x

What is the d/(dt)(((t+1)^2)/((t-1)^2)) ?
The d/(dt)(((t+1)^2)/((t-1)^2)) is -(4(t+1))/((t-1)^3)
What is the first d/(dt)(((t+1)^2)/((t-1)^2)) ?
The first d/(dt)(((t+1)^2)/((t-1)^2)) is -(4(t+1))/((t-1)^3)