In the solution to problem 1.1.11, there's is one line in the solution that is giving me trouble. Here is the problem:
Suppose that $$f$$ is a twice differentiable real function such that $$f''(x)>0$$ for all $$x\in[a,b]$$. Find all real numbers $$c$$ at which the area between the graph $$y=f(x)$$, the tangent to the graph at $$(c,f(c))$$, and the lines $$x=a$$, $$x=b$$ attains its minimum value.
At some point we need to differentiate (with respect to c)
$$A(c) = \int_a^b ( f(x)  f(c)  f'(c)(xc) ) dx$$
and get
$$A'(c) = f''(c) \int_a^b (xc) dx$$,
but I'm getting other things which don't allow successful solution to the problem (my expression for $$A'(c)$$ doesn't have a $$c$$ in it).
Can someone explain?
Derivative question from Berkeley Problems (Real Analysis)

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: Derivative question from Berkeley Problems (Real Analysis)
I don't understand the issue... Are you sure you're differentiating with respect to c...? Looks straightforward to me
Re: Derivative question from Berkeley Problems (Real Analysis)
First off, please note that your parentheses are off. You probably want to start with a function more like
$$A(x) = \int_a^b f(x)  f(c)  f'(c) \cdot (xc) \, dx$$
(I am not vouching for the correctness of the signs.) Using the chain rule to differentiate with respect to c yields
$$A'(x) = \int_a^b 0  f'(c)  \left( f''(c) \cdot (xc) + f'(c) \cdot (1) \right) \, dx$$
which simplifies to
$$A'(x) = f''(c) \int_a^b (xc) \, dx$$
$$A(x) = \int_a^b f(x)  f(c)  f'(c) \cdot (xc) \, dx$$
(I am not vouching for the correctness of the signs.) Using the chain rule to differentiate with respect to c yields
$$A'(x) = \int_a^b 0  f'(c)  \left( f''(c) \cdot (xc) + f'(c) \cdot (1) \right) \, dx$$
which simplifies to
$$A'(x) = f''(c) \int_a^b (xc) \, dx$$

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: Derivative question from Berkeley Problems (Real Analysis)
The steps look good owlpride, and although the parentheses look awkward, that's exactly what he had written above (there was just an extra set at the beginning and end of the integrand). Also I think you meant to call it A(c) instead of A(x) but that was just a typo. I guess it's also worth stating that you technically need a weak version of Leibniz in order to take the derivative underneath the integral sign.
Re: Derivative question from Berkeley Problems (Real Analysis)
My error was in treating $$f(c)$$ as a constant, "simplifying" the integrals, and differentiating that. Ah, the things you learn. : /
I'm now familiarizing myself with http://en.wikipedia.org/wiki/Differenti ... egral_sign . Thank you very much for the input.
No kidding Terry! http://terrytao.wordpress.com/careerad ... ourfield/
I'm now familiarizing myself with http://en.wikipedia.org/wiki/Differenti ... egral_sign . Thank you very much for the input.
No kidding Terry! http://terrytao.wordpress.com/careerad ... ourfield/
Re: Derivative question from Berkeley Problems (Real Analysis)
blitzer6266, you are completely right on all three points, thanks!