Using the chain rule on a fraction?

[LilTaru]

Homework Statement

Find dy/dx at x = 2.

y = (1 + s)/(1 - s); s = t - 1/t; t = sqrt(x)

Homework Equations

I know if f = f(g(x)), then f¹(x) = f¹(g(x)) * g¹(x)

The Attempt at a Solution

I think I may need to combine the chain rule and quotient rule, but all of the separate equations makes it really hard to keep track of all the steps. I have no idea where to even begin! Please help!

 

[Mentallic]

The chain rule can also be denoted as \frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx} given y=f(u) and u=f(x)

This can obviously be extended to 3 or more variables.

 

[LilTaru]

I think that is the part that confuses me... would dy/dx in this case be:
dy/ds * ds/dt * dt/dx?

 

[Mentallic]

Yep, it makes sense doesn't it?

If you look at them as being fractions, then of course \frac{dy}{dx}=\frac{dy}{ds}.\frac{ds}{dt}.\frac{dt}{dx} since the ds and dt's cancel from the numerator and denominator leaving dy/dx.
The second reason why it should make sense is because since we have y=f(s), s=f(t), t=f(x) then we can only take derivatives as dy/ds, ds/dt, dt/dx respectively.

 

[LilTaru]

Okay I think I may know what to do, but I am still a bit confused...

Would dy/ds * ds/dt * dt/dx be:

(1 - s)(s) - (1 + s)(-s)/(1 - s)² * t + 1/t² * 1/(2sqrt(x))

Since dy/ds = (1 - s)(s) - (1 + s)(-s)/(1 - s)²

ds/dt = t + 1/t²

and dt/dx = 1/(2sqrt(x))

Would I then just plug in for s, and t?

 

[Mentallic]

dy/ds and ds/dt are both wrong. You need to go back and revise your derivatives since this problem will keep creeping up on you and you'll never make it through the course.

 

[LilTaru]

Yeah that is the trouble I always have with these questions... how to differentiate it when it contains s and t... I was thinking of plugging in the values of s and t, but wouldn't that defeat the purpose of having them in the first place because I could just plug in the values and use the quotient rule?

 

[LilTaru]

I tried again and got:

dy/ds = 2/(1 - s)²

ds/dt = 1 + 1/t²

dt/dx = 1/(2sqrt(x))

I don't know if I am still really confused on it, but this seemed more correct to me than my first try... Do I now plug in for s and t or am I still really off base?

 

[Mentallic]

Yeah that is the trouble I always have with these questions... how to differentiate it when it contains s and t... I was thinking of plugging in the values of s and t, but wouldn't that defeat the purpose of having them in the first place because I could just plug in the values and use the quotient rule?

Yes you can do that, but trust me, you don't want to. It will become way too confusing and difficult to take the derivative of. Give it a shot if you want though, just to show yourself how ugly it will look and appreciate this approach much more

I tried again and got:

dy/ds = 2/(1 - s)²

ds/dt = 1 + 1/t²

dt/dx = 1/(2sqrt(x))

I don't know if I am still really confused on it, but this seemed more correct to me than my first try... Do I now plug in for s and t or am I still really off base?

Yes that's correct. Ok now since you're asked to find dy/dx at x=2, you know that since \frac{dt}{dx}=\frac{1}{2\sqrt{x}} then t=... and then since you now know t, then s=...? etc.

Of course using the chain rule formula that we already spoke about.

 

[Mentallic]

Yeah that is the trouble I always have with these questions... how to differentiate it when it contains s and t... I was thinking of plugging in the values of s and t, but wouldn't that defeat the purpose of having them in the first place because I could just plug in the values and use the quotient rule?

Yes you can do that, but trust me, you don't want to. It will become way too confusing and difficult to take the derivative of. Give it a shot if you want though, just to show yourself how ugly it will look and appreciate this approach much more

I tried again and got:

dy/ds = 2/(1 - s)²

ds/dt = 1 + 1/t²

dt/dx = 1/(2sqrt(x))

I don't know if I am still really confused on it, but this seemed more correct to me than my first try... Do I now plug in for s and t or am I still really off base?

Yes that's correct. Ok now since you're asked to find dy/dx at x=2, you know that since dt/dx=\frac{1}{2\sqrt{x}} then dt/dx=... and then since you now know t, then ds/dt=...? etc.

Of course using the chain rule formula that we already spoke about.

 

[LilTaru]

Okay so to clarify:

Then dt/dx = 1/(2sqrt(2)) = 1/2sqrt(2)

ds/dt = 1 + 1/(sqrt(2))² = 3/2

dy/ds = 2/(1 - sqrt(2) + 1/sqrt(2))² = 4/(2 - 2sqrt(2) + 1)

Then just multiply these together for the answer?

 

[Mentallic]

Yes that's perfect!

Just as a little reminder, sometimes you can clear things up a bit, such as the \sqrt{2}-\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}} but it's nothing really that important here.

Good job by the way.

 

[LilTaru]

Whew! Thank you so much! It really helps! Derivatives always get confusing when they start getting big and complicated! Your help was much appreciated!