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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x \to - 1 lim (\frac{x ^{2} - 2}{1 - x})

y^{^{'}} = 4 y^{2}

\frac{\partial}{\partial y} (600 - 4 x + 10 x^{2} y)

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

\int_{\frac{1}{2}}^{2} \frac{2}{x} - \frac{1}{2} x d x

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