# Calculus practice problem

My teacher assigned each student a practice problem yesterday for the new section we are starting, but I was absent, so I missed his explanation of how to do the problems. Can anyone explain to me how to solve it and provide the answer so I can practice it?

The problem is Suppose that $F(x)=f(\sqrt{1+x^2})$ such that $f'(\sqrt2)=\sqrt2/2$. Compute $F'(1)$.

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Please check that I interpreted everything correctly. Also, did you mean $F'(1)$ at the end? – Brian M. Scott Mar 8 '13 at 2:11
What you missed is pretty fundamental. It is called the chain rule: see here. – 1015 Mar 8 '13 at 2:12
Yes, I have to compute F'(1). But I missed the explanation of the "chain rule". I'm notoriously bad at math. – sucksatmath Mar 8 '13 at 2:24
In addition to the wiki link julien gave, I highly suggest looking at Khan Academy's videos on the chain rule. They're pretty good, and you should be able to do this problem after watching them. If you still need help, let us know. – apnorton Mar 8 '13 at 2:25

The chain rule states that $$\frac{d}{dx} f(g(x)) = g'(x)f'(g(x)).$$

Applying this here, we have

$$F'(x) = [\frac{d}{dx} \sqrt{x^2+1}] [f'(\sqrt{x^2+1})] = \frac{x}{\sqrt{x^2+1}} f'(\sqrt{x^2+1}).$$

Using $x=1$, we have

$$F'(1) = \frac{1}{\sqrt{2}} f'(\sqrt{2}) = \frac{1}{\sqrt{2}} \frac{\sqrt{2}}{2} = \frac{1}{2}.$$

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You guys are great here! Thanks so much! – sucksatmath Mar 8 '13 at 2:34

By the chain rule, $$F'(x)=f'(\sqrt{x^2+1})\frac{d}{dx}\sqrt{x^2+1}.$$ By the chain rule again, $$\frac{d}{dx}\sqrt{x^2+1}=\frac{1}{2}(x^2+1)^{-1/2}2x=\frac{x}{\sqrt{x^2+1}}.$$ So $$F'(x)=f'(\sqrt{x^2+1})\cdot\frac{x}{\sqrt{x^2+1}}.$$ Now make $x=1$.

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