Actually, looking back at the problem again, I needed to show that f'(r) > 0 implies g'(r) < 0, so I don't need to prove the converse. So, problem solved. Thanks for your help. I was tearing my hair out over that one.
It is desired to show that if, for constant k,
\frac{d}{dr} g(r) = \frac{d}{dr}k\frac{r^2}{\int^{r}_{0} tf(t)dt} < 0
Then
\frac{d}{dr}f(r) > 0
When I take the derivative of g, I get
\frac{dg}{dr} = k\frac{2r\int^{r}_{0}tf(t)dt - r^2(rf(r))}{(\int^{r}_{0} tf(t)dt)^2}
If that is correct, I...
That would be the problem. You didn't distribute the power correctly. What you did was
(x^2 + 5)^3 = (x^2)^3 + 5^3
But that's not true. The correct way to expand the power is
(x^2 + 5)(x^2 + 5)(x^2 + 5)
So, for example
(x+1)^2 = (x+1)(x+1) = x*x + 1*x + 1*x + 1*1 = x^2 + 2x + 1
Not...
Not quite. Remember that the chain rule says
\frac{d}{dx}f(g(x))=f'(g(x))g'(x)
When you have
f(x^2 - 1)
They're saying that, rather than plugging in the usual x in your variable, you plug in x^2 - 1. For example, let f(t) = t^2. Then, f(x^2) = (x^2)^2 = t^4. So, you have a...
You're very close. However, you forgot to apply the chain rule when you took the derivative of f(x^2 - 1). That, too, is a composition of function and so you must also apply the chain rule when taking its derivative.
I have a function of the form
\int^{r}_{0} tf(t)dt
I'm supposed to take the derivative with respect to r of this integral. By the fundamental theorem of calculus, is the derivative not
rf(r)
The problem being I need, for the problem to work out correctly, to have a df/dr term. So, am...