BuBbLeS01 said: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.
BuBbLeS01 said:Yes I got...
-1 * (1-t^2)^-1/2 + (2t)
(cos) * 1/2 * (1-t^2)^-3/2 + (2t)
real10 said:no that is incorrect... chain rule will be sin u *du where u in this case arcos(t)..
try again...
Dick said: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.
Yes that's right...where did cos go?real10 said:hmm I am getting [tex]\frac{-x}{(1-x^2)^(1/2)}[/tex] 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!
Is this right?BuBbLeS01 said:Yes that's right...where did cos go?
BuBbLeS01 said:cos (arccos t) * (-1/sqrt 1-t^2)
No, it simplifies to..t, not 1!BuBbLeS01 said: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
Oh yay! thanks...lolDick said:That's correct. If your teacher doesn't want you to simplify, I guess you can stop there.
Okay I got another one :)...I am getting a different answer then calcchat because I did it differently...BuBbLeS01 said:Oh yay! thanks...lol
The derivative of a function is the rate of change of the function with respect to its independent variable. In this case, the derivative of h(t) = sin(arccos(t)) is [-sin(arccos(t))] * [-1/sqrt(1-t^2)] or simply cos(arccos(t)) * 1/sqrt(1-t^2).
To find the derivative of a trigonometric function, you can use the chain rule, product rule, or quotient rule depending on the specific function. In this case, we use the chain rule to find the derivative of h(t) = sin(arccos(t)).
The chain rule is a method used to find the derivative of a composite function. It states that the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.
Yes, the derivative of h(t) = sin(arccos(t)) can be simplified to cos(arccos(t)) * 1/sqrt(1-t^2). Using the identity sin(arccos(t)) = sqrt(1-t^2), we can further simplify the derivative to 1/sqrt(1-t^2).
The derivative of a function represents the instantaneous rate of change of the function. In real-world applications, this can be used to model and analyze various phenomena, such as the growth or decay of a population or the speed of an object. In this specific case, the derivative of h(t) = sin(arccos(t)) can be used to find the instantaneous velocity of an object moving along a circular path at a given time t.