Calculus Help: Derivatives, Inverse Functions, and Tangent Lines Explained

[aizeltine]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

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

 

[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?

 

[aizeltine]

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?

I already answered number 1 so, don't 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 I am stuck
so i get 9y-6x^2 =0 then how do i get x?

 

[Dick]

I already answered number 1 so, don't 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 I am stuck
so i get 9y-6x^2 =0 then how do i get x?

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.

 

[aizeltine]

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?

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.

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

 

[Dick]

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

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

 

[aizeltine]

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

Number 2: I still don't get it?!?I don't know how to inverse f(1)
Number 3: I am 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?

 

[Dick]

Number 2: I still don't get it?!?I don't know how to inverse f(1)
Number 3: I am 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?

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