# Differentiation-dont really get the question

1. Jul 20, 2010

### pat666

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

y= sin(2t)+t
x=e2t
ln(x)=2t
t=ln(x)/2
y=sin(2(lnx/2) )+ln(x)/2
y=sin(ln(x))+ln(x)/2

the question says leave in terms of t but when i differentiate i get
y'=cos(ln(x))/x+1/2x

this may not even be what the question is asking??? but if it is how do i put that back in terms of t???

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2. Jul 20, 2010

### hunt_mat

What you have done is perfectly correct, but you haven't answered the question. Use the chain rule:
$$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$$
and the fact that $$dt/dx=1/(dx/dt)$$ and leave your answer in terms of t, just like it said. You have answered the harder question, well done.

3. Jul 20, 2010

### pat666

sorry I dont understand that, how do I leave it in terms of t??????

4. Jul 20, 2010

### hunt_mat

calculate dy/dt and dx/dt, these will be functions of t divide dy/dt by dx/dt, this will also be a function of t, is is this function they're asking for. You have done 99% of the problem already.

5. Jul 20, 2010

### pat666

what would dx/dt be????

6. Jul 20, 2010

### hunt_mat

you have the following:
y= sin(2t)+t
x=e^(2t)
so what is dy/dt and dx/dt? It's just simple differentiation.

7. Jul 20, 2010

### pat666

are ok thats simple
dx/dt=2e^(2t)
dy/dt=2cos(2t)+1
then I multiply them??

8. Jul 20, 2010

### hunt_mat

No, you divide them in a previous post I wrote:
$$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$$
How do you get from dx/dt to dt/dx?

9. Jul 20, 2010

### HallsofIvy

Staff Emeritus
Equivalently
$$\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
You get hunt mat's formula by "invert and multiply".

10. Jul 20, 2010

### pat666

so i get 2cos(2t)+1 * (2e^(2t))^-1
is that the right answer??? my 89 wont simplify it at all.

11. Jul 20, 2010

### hunt_mat

probably because you can't. It is correct, you can check it with your previous answer, find t in terms of x(you already did this), put it into the answer above and check that they're the same.

12. Jul 20, 2010