Derivative trigonometric functions help

[ada0713]

Derivative trigonometric functions

Homework Statement

Find d$$^{2}$$x/dt$$^{2}$$ as a function of x if dx/dt=xsinx

Homework Equations

The Attempt at a Solution

I tried to solve the problem by taking a second derivative of dx/dt=xsinx
but I was not sure how to start. Also, it is confusing because
doesn't have to be dx/dt=tsint in order to make sense (since i should
take the derivatice w/ respect to t, not x)?

I'm totally lost here, please help me with the start,
and i'll try to figure out the rest!

 

[HallsofIvy]

Homework Statement

Find d$$^{2}$$x/dt$$^{2}$$ as a function of x if dx/dt=xsinx

Homework Equations

The Attempt at a Solution

I tried to solve the problem by taking a second derivative of dx/dt=xsinx
but I was not sure how to start. Also, it is confusing because
doesn't have to be dx/dt=tsint in order to make sense (since i should
take the derivatice w/ respect to t, not x)?

I'm totally lost here, please help me with the start,
and i'll try to figure out the rest!

Fortunately your question "doesn't have to be dx/dt=tsint in order to make sense?" made me look at the question again! I had thought you were just asking for the second derivative of x sin(x) and obviously you aren't.

Yes, you need to differentiate with respect to t, not x. But dx/dt does NOT have to be an explicit function of x. Since x itself is a function of t, x sin(x) is an implicit function of t. If you are told that dx/dt= f(x), use the chain rule:
$$\frac{d^2 x}{dt^2}= \frac{df}{dt}= \frac{df}{dx}\frac{dx}{dt}$$
Just like "implicit differentiation" except this time you already know what dx/dt is.

 

[ada0713]

so..

This is what i did.
since x=f(t)
f'(t)=dx/dt
so, x'=dx/dt

if i take the second derivative of dy/dt,
it'll be like
(dy/dt)'= 1*(dx/dt)sinx + x*cosx* (dx/dt)

the red part is where i used the chain rule.
since I don't know the actual function of x, I just set the derivative of it as
dx/dt (because I'm taking derivative w/ respect to t)

and since this is basically the derivative of of product, i used
a product rule to do the rest.

I simplfied it and ended up with
dx/dt(sinx+xcosx)
am i right..? my answer looks quite a bit complicated,,

 

[Dick]

That looks right.