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...
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...
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...
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.