What is the derivative of f(t) when integrated with inverse trig functions?

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SUMMARY

The derivative of the function f(t), defined as the integral of 1/(1+x^2) from t² to 0, is f'(t) = -2t/(1 + t⁴). This result utilizes the integration of inverse trigonometric functions, specifically the arctangent function, which is integral to understanding the behavior of this function. The discussion highlights the importance of recognizing the limits of integration and their implications on the derivative calculation.

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If f(t)=integral of 1/(1+x^2) from t^2 to 0,then f'(t)=?

thanks very much.
 
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The integral looks odd since t^2 > 0 and you have 0 as the upper limit.. However as it stands, f'(t)=-2t/(1 + t^4)
 
Use integration of inverse trig functions.
 

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