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

The derivative of (t^2+1)/(t^2-1) is -(4t)/((t^2-1)^2)

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

The first derivative of (t^2+1)/(t^2-1) is -(4t)/((t^2-1)^2)

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derivative of

\frac{t ^{2} + 1}{t ^{2} - 1}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= - \frac{4 t}{( t ^{2} - 1 ) ^{2}}

t a y l or x e^{- x^{2}}

\frac{\partial}{\partial x} (e^{x + y + 3})

t an g e n t y = \frac{x}{81 + x ^{2}}, (0, 0)

y^{^{''}} - 4 y^{^{'}} - 5 y = 0

a re a x + y^{2} = 42, x = y

What is the derivative of (t^2+1)/(t^2-1) ?
The derivative of (t^2+1)/(t^2-1) is -(4t)/((t^2-1)^2)
What is the first derivative of (t^2+1)/(t^2-1) ?
The first derivative of (t^2+1)/(t^2-1) is -(4t)/((t^2-1)^2)