Recent content by Dustobusto

Yeah I'm starting to realize that wasn't right x3 - x6 = 3x3-3 is that at least on the right track?

OK, so x3-y3 = 3xy -3 Substitute, x3 -(x2)3 = 3(x)(x2) - 3 x3 - x6 = 3x3-3 Subtract 3x3 both sides and get -2x3-x6 = -3 So at this point, I'm not sure what's factorable and what's not. I could factor out an x3 and get x3(-2-x3) = -3 but I'm not sure how much that helps me. Do I need to...

Well I understand part of what you're saying. I can solve the y' so that x2 = y and then go back into the equation of y' and replace y with x2 and demonstrate that it equals zero. Do I plug x2 into the original y and solve for x? Or take the derivative and solve for x?

So in finding a pattern for the nth derivative for y = x-1 the book wrote y(n)(x) = (-1)nn!x-n-1 What does the exclamation mark mean? I know != means not equal to.

Homework Statement Find the points on the graph of x3-y3=3xy-3 where the tangent line is horizontal Homework Equations y = f(x) so implicit differentiation must be used when taking the derivative of y (xy)' = xy' + y The Attempt at a Solution So if the tangent line is...

ln(a * b) = ln(a) + ln(b) and ln(a/b) = ln(a) - ln(b) correct?

Okay. Also, with x2ex(x+3) = 0 x = -3 but does it also = 0? Graphing it, it looks like I could include that value as well.

Homework Statement Find the points on the graph of y = x3ex where the tangent line is horizontal. (Your answer should include x-values and y-values) Homework Equations Product rule fg'+gf' The Attempt at a Solution (x3)(ex)+(ex)(3x2) x3ex+3x2ex= y' x3ex+3x2ex= 0...

Homework Statement f(x) = x ln(x) The attempt at a solution f'(x) = product rule, resulting in 1 + ln (x) So by way of solving the problem, set 1 + ln (x) = 0 Now idealistically, find something in x that, when added to 1, equals zero. Now here's a problem I have. How do I know when to...

So I have an exam tomorrow, and the teacher provided a review. f(x) = ln(x + y) I remember that d/dx ln[f(x)] = f'(x)/f(x) so would that not equal 2/(x + y) ? The answer she gave is 1/(x + y - 1) ... where that neg. one came from I have no idea. Come to think of it, there were no...

Yeah, it's starting to dawn on me what this all means. I blame the book though :)

So, I understand that implicit differentiation involves derivatives in which x values and y values are mixed up. I've done several implicit differentiation problems a couple sections ago for my math homework, but I pretty just memorized patterns and solved it that way. Now that I'm trying to...

Yeah, I caught it just before you posted that lol. Thanks Mark.

Yes, it makes complete sense now. It ends up (2x-2)(ln 5)58 I believe you can write that out as 2x ln 5 * 2 ln 5 and, x is 1, so 2 ln 5 - 2 ln 5 = 0 Zero times 58 is zero. Got it.

Ok, I think I know at least the beginning of where I went wrong. We now have (ln 5)5u(x2-2x+9)' and since x = 1 its (ln 5)5u * 8 which gives us (8 ln 5)58 The derivative of (x2-2x+9) isn't 8. It's 2x-2. I think this is where I need to start.