# Question Need help

1. Oct 7, 2007

Derivative trigonometric functions

1. The problem statement, all variables and given/known data
Find d$$^{2}$$x/dt$$^{2}$$ as a function of x if dx/dt=xsinx

2. Relevant equations

3. 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)?

and i'll try to figure out the rest!!

Last edited: Oct 7, 2007
2. Oct 7, 2007

### HallsofIvy

Staff Emeritus
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.

3. Oct 7, 2007

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 fucntion 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,,

4. Oct 7, 2007

### Dick

That looks right.