Derivative trigonometric functions help

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The discussion revolves around finding the second derivative of x with respect to t, given that dx/dt = x sin(x). Participants clarify that differentiation should be performed with respect to t, not x, and that dx/dt does not need to be an explicit function of x. The chain rule is emphasized for deriving the second derivative, leading to the expression d^2x/dt^2 = dx/dt(sin(x) + x cos(x)). The solution process involves using both the chain rule and product rule for differentiation. The final expression is confirmed to be correct, despite its complexity.
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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!
 
Last edited:
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ada0713 said:

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.
 
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,,
 
That looks right.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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