# 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)