Calculus HELP!
find the derivative of the function.
g(t)=5cos^3(times pie)(times t)

- anonymous

Calculus HELP!
find the derivative of the function.
g(t)=5cos^3(times pie)(times t)

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- zepp

\(\large g(t) = 5cos^3(\pi t)\)

- ksaimouli

HINT - pie and t are constant

- ksaimouli

use the trig formulas for derivatives

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## More answers

- anonymous

ok zepp so tell me if i have it right so far.....
y(prime)=u(prime) times v+v(prime) times u
= no wait.... im stuck

- anonymous

where do i take the derivative first for this function?

- zepdrix

Take the derivative of the OUTERMOST function first. Trig functions can be a little tricky to read since the power is written in a different location.
Identify the outermost function.
\[5(\cos(\pi t))^3\]
That is another way it can be written so you can easily identify the outermost function.

- UnkleRhaukus

use the chain rule , not the product rule

- anonymous

ok so whta rule would i be using in this case when i take ther derivative of the outermost function

- zepdrix

The power rule :) and then as Unkle mentioned, the chain rule after that.

- anonymous

ok..... so using the power rule in order to find the outermost function first this would be 15?

- zepdrix

\[15(\cos(\pi t))^2 * (d/dt)(\cos(\pi t))\]
good you got the first step :) by the chain rule, you must now multiply by the derivative of the inside, as I have written here.

- UnkleRhaukus

\[(f\circ h)'(x)=f(h(x))'=f'(h(x))\cdot h'(x)\]

- anonymous

ok i got this step now i must find the derivative of (cos(piet))^2 right?

- zepdrix

Don't look at the square anymore. You already dealt with that outermost function. the inner function is cos(pi t)

- anonymous

ok so i find the derivative of that and that is -sin(piet) right?

- zepdrix

\[15(\cos(\pi t))^2*(-\sin(\pi t))*(d/dt)(\pi t)\]
Good! now we have one last step. there is an inner inner function that you need to differentiate still, as I have written above.

- anonymous

ok! thanks! ummm... would this be no wait u lost me

- zepdrix

The innermost function is (pi t). So you need to multiply by the derivative of that :D Find the derivative! :D

- anonymous

u see i thought of that being the answer but i wasn't quite sure... umm okay so the derivative of piet is just piet?

- zepdrix

\[(d/dt)(\pi t)=\pi(d/dt)t=?\]

- zepp

\(\pi\) is a constant, and when a constant combined to a variable, the variable goes to 1 but the constant stay.

- anonymous

zepdrix :[

- anonymous

im sorry u guys lost me completely

- zepp

Let's restart everything, haha

- zepdrix

nooo don't get lost :D you were so close!

- anonymous

is it just pie?

- zepdrix

yayyyy \:D/

- zepdrix

pi* lol

- UnkleRhaukus

\[g(t) = 5\cos^3(\pi t)\]
\[f(x)=5\cos^3(x)\]
\[h(x)= \pi x\]

- anonymous

CHAIN RULE :D

- zepp

Okay, \[\large g(t) = 5cos^3(\pi t)\]We identified \(5cos(\pi t)\) as the inner function, and \((Stuffshere)^3\) as the outter function, so we could apply the power rule here and we apply the chain rule
\[3*5(cos^2(\pi t))*\frac{d}{dt}cos^2(\pi t)*\frac{d}{dt}(\pi t)\]

- anonymous

ok how it pi though

- anonymous

how is it pi

- zepp

\(\large 3*5(cos^2(\pi t))*\frac{d}{dt}cos(\pi t)*\frac{d}{dt}(\pi t)\)***
Sorry

- zepdrix

what is the derivative of 2x? its just 2 times the derivative of x.
So you apply the power rule, (derivative of x^1 = 1.. so you end up with 2. :O yes?
same thing is happening here. don't let the pi confuse you, its just a number

- anonymous

ok !!!! i think i got!!! THANKS SOOOOO MUCH I WAS HAVING TROUBLEE!!!!!!

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