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

[BuBbLeS01]

Homework Statement

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.

 

[real10]

Homework Statement

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.

did u try chain rule?

 

[BuBbLeS01]

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

 

[real10]

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

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

 

[Dick]

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

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.

 

[real10]

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.

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!

 

[Dick]

I'll let BuBbLeS01 check that.

 

[BuBbLeS01]

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!

Yes that's right...where did cos go?

 

[BuBbLeS01]

Yes that's right...where did cos go?

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

 

[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.

 

[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}}}

 

[BuBbLeS01]

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

 

[arildno]

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

Correct!

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

 

[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

 

[arildno]

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

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

 

[BuBbLeS01]

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

 

[Dick]

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

 

[BuBbLeS01]

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

 

[Dick]

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

 

[BuBbLeS01]

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

 

[Dick]

That's correct. If your teacher doesn't want you to simplify, I guess you can stop there.

 

[BuBbLeS01]

That's correct. If your teacher doesn't want you to simplify, I guess you can stop there.

Oh yay! thanks...lol

 

[BuBbLeS01]

Oh yay! thanks...lol

Okay I got another one :)...I am getting a different answer then calcchat because I did it differently...
Find the derivative...
y = xarcsinx + sqrt 1-x^2
y = xarcsinx + (1-x^2)^1/2
y' = x(1/sqrt1-x^2) + 1/2(1-x^2)^-1/2 * (2x)