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

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

x \to 2 lim ((2 x + 5)^{x - 3})

\frac{6 x}{x + 5}, at x = - 3

\int_{- 1}^{1} 0 d x

n = 0 \sum \infty \frac{8}{5 ^{n}}

(x^{3} + 9) e^{x}

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