1. ## Implicit differentiation - Help please!

Probably simple, but it's not exactly apparent to me...

1. y^3 - x^2 -1 =x^2 - 3x

A. Find dy/dx. **I came up with y'= 4x-3 / 3y^2 for this part.

B. Use algebra to solve for y. **I said y= (2x^2 - 3x + 1)^(1/3)

C. Use "traditional methods" to differentiate y from part B.
**WHAT? Do I use the quadratic formula here?

D. Show that the answers in part A and C are the same using algebra.
**Maybe after I figure out what to do with part C,
this will become clear.

And now for what I think is probably an easy one:

2. A particle is moving along the graph of the function y= x^2 + 3x - 5 at a rate of dx/dt = 3cm/s. Find the rate of change of the particle in the y direction dy/dt, when x=1.

**Do I take the natural log of each side first? Then, solve for y', plug in x=1? What do I do with the 3cm/s?

Hopefully someone can give me a jump start. Thanks.

2. Originally Posted by Esiuol
Probably simple, but it's not exactly apparent to me...

1. y^3 - x^2 -1 =x^2 - 3x

A. Find dy/dx. **I came up with y'= 4x-3 / 3y^2 for this part.
corerct

B. Use algebra to solve for y. **I said y= (2x^2 - 3x + 1)^(1/3)
correct

C. Use "traditional methods" to differentiate y from part B.
**WHAT? Do I use the quadratic formula here?
"what" is right! what does the quadratic formula have to do with differentiation. you somehow got amazingly confused here, after doing so well up to this point. use the chain rule

D. Show that the answers in part A and C are the same using algebra.
**Maybe after I figure out what to do with part C,
this will become clear.
yes, figure out C first

3. Originally Posted by Esiuol

2. A particle is moving along the graph of the function y= x^2 + 3x - 5 at a rate of dx/dt = 3cm/s. Find the rate of change of the particle in the y direction dy/dt, when x=1.

**Do I take the natural log of each side first? Then, solve for y', plug in x=1? What do I do with the 3cm/s?
explain your logic, why would you take the log of both sides?

$y = x^2 + 3x - 5$

differentiating implicitly we get:

$\frac {dy}{dt} = 2x~\frac {dx}{dt} + 3~\frac {dx}{dt}$

now just plug in the values you were given and solve for what you want

4. Ooooooh, ok...the "traditional methods" part tripped me up. I thought he was asking me not to use calculus or something.

Any thoughts on that second one?

5. Thanks!!!