Derivative of function h(t) = sin(arccos(t)) - help

1. Oct 4, 2007

BuBbLeS01

1. The problem statement, all variables and given/known data
find the derivative of the function:
h(t) = sin (arccos t)

2. The attempt at a solution
derivative of sin = cos
derivative of arccos t:
(-1/sqrt 1-t^2) * 1

my book has a complete different answer so I need some help please.

2. Oct 4, 2007

real10

did u try chain rule?

3. Oct 4, 2007

BuBbLeS01

Yes I got...
-1 * (1-t^2)^-1/2 + (2t)
(cos) * 1/2 * (1-t^2)^-3/2 + (2t)

4. Oct 4, 2007

real10

no that is incorrect... chain rule will be cos u *du where u in this case arcos(t)..
try again...

Last edited: Oct 4, 2007
5. Oct 4, 2007

Dick

No, it won't. BuBbLeS01's first solution was much closer. Derivative of sin(f(t)) is cos(f(t))*f'(t). Let f be arccos.

6. Oct 4, 2007

real10

hmm I am getting $$\frac{-x}{(1-x^2)^(1/2)}$$ which I believe is right unless i read the question incorrectly!..
aaah a typo I have corrected the typo should be cos.thanks for pointing that out!

Last edited: Oct 4, 2007
7. Oct 4, 2007

Dick

I'll let BuBbLeS01 check that.

8. Oct 5, 2007

BuBbLeS01

Yes thats right...where did cos go?

9. Oct 5, 2007

BuBbLeS01

Is this right?
cos (arccos t) * cos(-1/sqrt 1-t^2) * (2t)
My teacher does not want our answers simplified at all.

10. Oct 5, 2007

Dick

You are mixing up arccos and it's derivative. cos(arccos(t))=t. cos and arccos are inverse functions, right? The derivative of sin(f(t))=cos(f(t))*f'(t). That's the chain rule. Now carefully put f=arccos.

11. Oct 5, 2007

arildno

Okay, we have:
1. $$h(t)=\sin(arccos(t))$$

2. $$\frac{d}{dx}\sin(x)=\cos(x)$$

3. $$\frac{d}{dt}arccos(t)=\frac{-1}{\sqrt{1-t^{2}}}$$

4. $$\cos(arccos(t))=t$$

Agreed?

Thus, by the chainrule, setting $x(t)=arccos(t)$, we have:
$$\frac{dh}{dt}=\cos(x(t))\frac{dx}{dt}=-\cos(arccos(t))\frac{1}{\sqrt{1-t^{2}}}=-\frac{t}{\sqrt{1-t^{2}}}$$

12. Oct 5, 2007

BuBbLeS01

cos (arccos t) * (-1/sqrt 1-t^2)

13. Oct 5, 2007

arildno

Correct!

Now, what does cos(arcos(t)) simplify to?

14. Oct 5, 2007

BuBbLeS01

Ok so cos(arccos(t)) simplifies to 1 because they are inverses of each other. I am sorry it has been a while since I have had precalc so I rusty

15. Oct 5, 2007

arildno

No, it simplifies to..t, not 1!
They are functional inverses, not multiplicative inverses.

16. Oct 5, 2007

BuBbLeS01

ok so anytime I have like sin(arcsin(t)) it just cancels to t?

17. Oct 5, 2007

Dick

Sure, just like (sqrt(t))^2=t.

18. Oct 5, 2007

BuBbLeS01

Okay I have another similar problem...
Find the derivative of...
tan(arcsint)
can I use the product rule?

19. Oct 5, 2007

Dick

It's not a product. Use the chain rule.

20. Oct 5, 2007

BuBbLeS01

sec^2(arcsint) * (1/sqrt1 + t^2)