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I have the find the derivative of the following function: $$F(t) = \int_1^{t^2} \frac{\sqrt{1+s^2}}{s} ds$$ If the upper limit of the integral was $t$ rather than $t^2$, this would be an easy application of the fundamental theorem of calculus. As it stands, I don't really know what to do with it. Help?

Jacob
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    Use the chain rule. –  Dec 15 '14 at 02:55
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    Note that $F=g\circ h$, where $h(t)=t^2$ and $\displaystyle g(t)=\int \limits_1^{t} \dfrac{\sqrt{1+s^2}}{s} \mathrm ds$, for all $t\in [1,+\infty[$. – Git Gud Dec 15 '14 at 02:56
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    Here is a similar type of problem: http://math.stackexchange.com/questions/1047523/derivative-of-integral-with-x-as-the-lower-limit/1047529#1047529 – Cameron Williams Dec 15 '14 at 02:58

2 Answers2

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Let $G(s)$ be an anti-derivative of $$g(s)=\frac{\sqrt{1+s^2}}{s}.$$ By the Fundamental Theorem of Calculus, $$F(t)=\int_1^{t^2}g(s)ds=G(t^2)-G(1)$$ so $$F'(t)=2tg(t^2)=\frac{2\sqrt{1+t^4}}{t}.$$

Spenser
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$$F'(t)=\frac d{dt} \left(\int_1^{t^2} \frac{\sqrt{1+s^2}}{s} ds \right)=\frac{\sqrt{1+t^2}}{t} \cdot \frac d{dt}(t^2)$$

Cookie
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