Solve Chain Rule Problem: Finding dy/dx for y=2t+3 and x=t^2-t

[Checkfate]

I need help solving this problem. It is in my textbook 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.

 

[arildno]

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

 

[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!

 

[chaoseverlasting]

dy/dt =2

dx/dt=2t-1

dt/dx= 1/2t-1

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

 

[Checkfate]

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

 

[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.

 

[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.