1. Sep 19, 2006

### Checkfate

I need help solving this problem. It is in my text book but no answer is provided in the appendix.

If $$y=2t+3$$ and $$x=t^{2}-t$$, find $$\frac{dy}{dx}$$

In theory this should be fairly straight forward! Simply find $$\frac{dy}{dt}$$ and $$\frac{dt}{dx}$$ and multiply both derivatives together to find $$\frac{dy}{dx}$$ , but I am having some problems.

I tried solving $$x=t^{2}-t$$ for t, but that gets ugly pretty fast. I am out of ideas, can someone point me in the right direction? Thanks again in advance.

2. Sep 19, 2006

### arildno

Remember the rule for the derivative of the inverse function..

3. Sep 19, 2006

### Checkfate

Sorry

I am sorry but I have never heard of that rule. I just looked through my textbook and I cannot find any mention of it :( This is for grade 12 calculus. could you just elaborate a little bit? sorry and thanks!

4. Sep 19, 2006

### chaoseverlasting

dy/dt =2

dx/dt=2t-1

dt/dx= 1/2t-1

dy/dx=2/(2t-1)

5. Sep 19, 2006

### Checkfate

ahhhh that all makes perfect sense now!!! But wouldn't dt/dx = -1/2t-1... :)

6. Sep 19, 2006

### HallsofIvy

Parentheses!! Since x= t2- t dx/dt= 2t- 1 and so dt/dx= 1/(2t-1). I see no reason for a negative sign there.

7. Sep 20, 2006

### Checkfate

Yea I realized that a mintute after I said it.. No need for a perpendicular derivative! lol, don't know what I was thinking, but anyways I understand 100% now. Thanks! Just to make this clear, does this apply to alll instances where you have $$\frac{dy}{dx}$$ and want to find $$\frac{dx}{dy}$$ or vice versa? Thanks. I would imagine it would but I just want to make sure.