Differentiation-dont really get the question

[pat666]

Homework Statement

see attachment please

Homework Equations

The Attempt at a Solution

y= sin(2t)+t
x=e^2t
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?

 

[hunt_mat]

What you have done is perfectly correct, but you haven't answered the question. Use the chain rule:
<br /> \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}<br />
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.

 

[pat666]

sorry I don't understand that, how do I leave it in terms of t?

 

[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.

 

[pat666]

what would dx/dt be?

 

[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.

 

[pat666]

are ok that's simple
dx/dt=2e^(2t)
dy/dt=2cos(2t)+1
then I multiply them??

 

[hunt_mat]

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

 

[HallsofIvy]

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

 

[pat666]

so i get 2cos(2t)+1 * (2e^(2t))^-1
is that the right answer? my 89 won't simplify it at all.

 

[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.

 

[pat666]

ok thanks for your help