Help in Calculus please?

1. Dec 21, 2011

aizeltine

1. The problem statement, all variables and given/known data
Can you please explain and show the steps of the ff.?
1. Find the derivative of the ff. f(x) = cos^5(2x^6)
2. Suppose f(1) = 2, f '(1) = 3 and f '(2) =4, then f^-1(2)=?

3. Find all points on the curve 2x^3 + 2y^3 -9xy= 0 where you will have a horizontal and vertical tangent lines.

2. Relevant equations

chain rule, i guess, and then the dx/dy thing if you have a y variable

3. The attempt at a solution
1. f'(x)= -sin(2x^6)* 2 ( but its not one of the choices)
2. I dont know what equation to use!
3. horixontal means slope is 0 and vertical means slope is undefine. I dont know what equation i need to set equal to zero and to undefine

2. Dec 21, 2011

e.bar.goum

Can you show the steps you used to get to question 1? Then we can tell you where you went wrong (because you did)

For 3, it's multivariable, so find points of 0 slope in x and in y - do you know how to implicitly differentiate?

3. Dec 21, 2011

aizeltine

I already answered number 1 so, dont worry about that, the only thing that I have problem with is 2 and 3
For number 2, where do I plug in those numbers?
For number 3, here's what I got.
6x^2+6y^2dy/dx - (9y +9x dy/dx)=0
6x^2 + 6y^2dy/dx - 9y -9x dy/dx=0
6y^2dy/dx-9xdy/dx=9y-6x^2
dy/dx= (9y-6x^2)/(6y^2-9x)
And then i set both numerator and denom equal to 0
but im stuck
so i get 9y-6x^2 =0 then how do i get x?

4. Dec 21, 2011

Dick

Solve 9y-6x^2 =0 for y and substitute it into original equation, 2x^3 + 2y^3 -9xy= 0. And for the second one, f(1)=2 alone should be enough to tell you what f^(-1)(2) is.

5. Dec 21, 2011

aizeltine

oh ok,,i get numberr 3 now, for number 2 should i take f(1) to the negative 1 power?

6. Dec 21, 2011

Dick

f^(-1) in this case doesn't mean 1/f(x), I don't think. It means the inverse function.

7. Dec 21, 2011

aizeltine

Number 2: I still dont get it?!?I dont know how to inverse f(1)
Number 3: Im stuck
so i got y = 2x^2/3 and I plugged that in into orig eqn and got
2x^3 + 2 (2x^2/3)^3-9x(2x^2/3)=0
2x^3+2(8x^6/27)-18x^3/3=0
2x^3+16x^6/27-6x^3=0
-4x^3+16x^6/27=0
(-108x^3+16x^6)/27=0
-108x^3+16x^6=0
4x^3(-27+4x^3)=0
x=0 and 3rdroot of 27/4
why do i have 2 x's???which one will i use?

8. Dec 21, 2011

Dick

It didn't say there was one point with a horizontal tangent. Maybe there are two. It said find all of them. And for the second one check out http://en.wikipedia.org/wiki/Inverse_function

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