Math Help - Derivatives derivatives...

1. Derivatives derivatives...

man i am just getting stump on every question of my practice exam...
heres where im getting lost

Complete the following

du/dv, where u=y^2 and v=y+3y^3 where do i start? take the derivative of each?

dy/dx where y=f(1-2x) and f ' (x)= sec (x)

dy/dx at the point (1,2), where y=y^2-2x^3

2. Originally Posted by cyberdx16
man i am just getting stump on every question of my practice exam...
heres where im getting lost

Complete the following

du/dv, where u=y^2 and v=y+3y^3 where do i start? take the derivative of each?
Paramentric curve.

Find the derivative of the first,
u'=2y
Now the derivative of the second,
v'=1+9y^2
Then their ratio is the derivative,
u'/v'=2y/(1+9y^2)

dy/dx where y=f(1-2x) and f ' (x)= sec (x)
Do not understand.

dy/dx at the point (1,2), where y=y^2-2x^3
Implictly:
y'=2yy'-6x^2
Substitute (1,2)
y'=2(2)y'-6(1)^2
Thus,
y'=4y'-6
Thus,
-3y'=-6
Thus,
y'=2

3. Originally Posted by ThePerfectHacker
Paramentric curve.

Find the derivative of the first,
u'=2y
Now the derivative of the second,
v'=1+9y^2
Then their ratio is the derivative,
u'/v'=2y/(1+9y^2)

Do not understand.

Implictly:
y'=2yy'-6x^2
Substitute (1,2)
y'=2(2)y'-6(1)^2
Thus,
y'=4y'-6
Thus,
-3y'=-6
Thus,
y'=2

how would u go about and find the second derivative? as well
how do u get y'=2yy'-6x^2, is because when dy/dx, which means to find y as the function of x? thx!

4. Originally Posted by cyberdx16
man i am just getting stump on every question of my practice exam...
heres where im getting lost

Complete the following

du/dv, where u=y^2 and v=y+3y^3 where do i start? take the derivative of each?
du/dv=du/dy dy/dv

Also when v is a well enough behaved function of y dy/dv=1/(dv/dy).

Now du/dy=2y, and dv/dy=1+9y^2, so:

du/dv=2y/(1+9y^2)

Now we can if we wish rewrite the right hand side in terma of u and v.

du/dv=2y/(1+9u),

and as

v=y+3y^3=y(1+3y^2)=y(1+3u)

we have:

y=v/(1+3u),

hence:

du/dv=2v/[(1+3u)(1+9u)]

5. Originally Posted by cyberdx16

dy/dx where y=f(1-2x) and f ' (x)= sec (x)
You need the chain rule here:

d/dx [f(g(x))] = f'(g(x))g'(x)

So here we have g(x)=1-2x, so g'=-2 so:

dy/dx = f'(1-2x)(-2) = -2 sec(1-2x)

RonL