? I don't understand this answer

[moe darklight]

?? I don't understand this answer

Homework Statement

The pdf attached is the answer to a question from a Calculus quiz. (the right hand side is a bit cut off, but the formula is repeated further down)

Edit: the "f(x) = ..., and a=0" part is the solution, not part of the the question.

Homework Equations

The Attempt at a Solution

ok. I have no idea what she's doing here. when I do the derivative of the f(x) she gives as the answer, I get

1/2 (1+3cosx)^1/2 is

1/4 ((1+3cosx)^-1/2)(-3sinx) = -3sinx / 4 sqrt(1+3cosx)

which is not even close to the original f'(x) she posted...

can anyone show me how she got that answer? thanks.

 

[HallsofIvy]

The derivative df/dx of f(x) at x= a is, by definition
$$\lim_{h\righrarrow 0}\frac{f(a+h)- f(a)}{h}$$

You are asked to find a function f and number a such that
[tex]\lim_{x\rightarrow 0}\frac{\sqrt{1+ 3cos(x)}- 2}{2x}= f'(a)[/itex]

Compare that to the definition of the derivative. If you replace the x with h then you have
[tex]\lim_{h\rightarrow 0}\frac{\sqrt{1+ 3cos(h)}- 2}{2h}= f'(a)[/itex]
or
[tex]\lim_{h\rightarrow 0}\frac{(1/2)\sqrt{1+ 3cos(h)}- 1}{h}= f'(a)[/itex]

Since $(1/2)\sqrt{1+ 3cos(0)}= (1/2)\sqrt{4}= 1$, it should be clear that you take f(x) to be $f(x)= (1/2)\sqrt{1+ 3cos(x)}$ and a= 0.

 

[moe darklight]

thanks. wow. I can't believe I didn't see that. gah. I did so poorly on that quiz. I don't know what happened. I was getting 80's and 90's, and on this quiz this morning I failed... and miserably. the whole thing was just a staring contest between me and the page. It's like I woke up 50% dumber today. I felt like writing an apology at the beginning to whoever marks it. This quiz is going to bring my overall mark down by 5%. bah.