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Popular >

((2t)/(t^2+1))^'

Solution

(\frac{2 t}{t ^{2} + 1})^{^{'}}

Solution

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

Solution steps

(\frac{2 t}{t ^{2} + 1})^{'}

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 (\frac{t}{t ^{2} + 1})^{^{'}}

Apply the Quotient Rule:

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

= 2 \cdot \frac{t ^{^{'}} ( t ^{2} + 1 ) - ( t ^{2} + 1 ) ^{^{'}} t}{( t ^{2} + 1 ) ^{2}}

t^{'} = 1

(t^{2} + 1)^{'} = 2 t

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

Simplify

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

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

Graph

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Frequently Asked Questions (FAQ)

What is the ((2t)/(t^2+1))^' ?
The ((2t)/(t^2+1))^' is (2(-t^2+1))/((t^2+1)^2)
What is the first ((2t)/(t^2+1))^' ?
The first ((2t)/(t^2+1))^' is (2(-t^2+1))/((t^2+1)^2)