Problem with Integrals - Finding f'(pi/2) in Calc 1

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SUMMARY

The discussion centers on finding the derivative f'(π/2) of the integral function f(x) defined by the integral of 1/(1+t^3)^(1/2) from 0 to g(x), where g(x) is the integral from 0 to cos(x) of (1+sin(t^2))dt. The correct approach involves applying the chain rule, where f'(x) is expressed as F'(g(x)) * g'(x), simplifying the evaluation at a single point. The user initially misinterpreted the relationship between f'(x) and g(x), but clarification was provided on the differentiation process.

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Homework Statement



Ok, I am given a integral f(x) with a lower limit 0 zero and an upper limit of g(x). Its 1/(1+t^3)^(1/2)dt

g(x) is the integral from 0 to cosx of (1+sin(t^2))dt. I need to find f '(pi/2). Any suggestions on where to start? This is calc one. I can find g '(x) but I don't think that's what is require. I am not sure however on how to actually integrate it.

f '(x) would be 1/(1+(g(x))^3) but I Dont know what to do with g(x).

Homework Equations



find f'(pi/2)

The Attempt at a Solution



f '(x) would be 1/(1+(g(x))^3) but I Dont know what to do with g(x).
 
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intervade said:
f '(x) would be 1/(1+(g(x))^3) but I Dont know what to do with g(x).

Not quite right. Denote the integral as F(g(x)) - F(0). Now, differentiating this, you get F'(g(x)) * g'(x). When you get to dealing with g(x), remember that you only need to evaluate it in a single point; this should make things a bit easier.
 

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