1. ## dy/dx for paramatic

Hey All,

find dy/dx

where y(t)=sin(t+7), x(t)=tln|t|

I'm having trouble with the |t| get abit confusing. hope someone can help. Thank you

Hey All,

find dy/dx

where y(t)=sin(t+7), x(t)=tln|t|

I'm having trouble with the |t| get abit confusing. hope someone can help. Thank you
d/dt(|t|) = {1 for t >= 0 and -1 for t < 0}, so you need to split your real line into intervals.

dx/dt = ln|t| + t*(1/t)*1

-Dan

(Corrected version! Sorry about the mistake.)

3. ## re:

are you applying the chain rule?

sorry im still abit confused

4. ## Re:

RE:

No for

x(t)=tln|t| you would use the product rule....

f'(x)*g(x)+f(x)*g'(x)

5. Originally Posted by qbkr21
RE:

No for

x(t)=tln|t| you would use the product rule....

f'(x)*g(x)+f(x)*g'(x)
i think he was talking about using the chain rule to find dy/dx. if so, then the answer is yes. you would use dy/dx = dy/dt * dt/dx

6. I disagree with Dan’s derivative of (t)[ln(|t|)], maybe he misread it.
Yes, we must use the chain rule.

7. Originally Posted by topsquark
d/dt(|t|) = {1 for t >= 0 and -1 for t < 0}, so you need to split your real line into intervals.
Or you can use the fact that,

(|x|) ' = sgn (x) for x!=0

8. Originally Posted by ThePerfectHacker
Or you can use the fact that,
(|x|) ' = sgn (x) for x!=0
Yes, that is true enough.
But the point is: [ln(|x|)]'=(1/x), x!=0.

9. thank you thats great guys

10. Are you sure? If they go un-interrupted, I bet the people on this forum could go on for another 12 pages on how to take the derivative of |x|.

11. Originally Posted by ecMathGeek
Are you sure? If they go un-interrupted, I bet the people on this forum could go on for another 12 pages on how to take the derivative of |x|.
lol

yeah i was thinking this question got quite a good response.

12. And if you don't see right away why the derivative of |x| would be sgn(x) (for x nonzero), rewrite |x| = sqrt(x²). Now take the derivative:

1/(2.sqrt(x²) * 2x = x/sqrt(x²) = x/|x| = sgn(x), by definition.