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lilly21
Calculus HELP! find the derivative of the function. g(t)=5cos^3(times pie)(times t)
\(\large g(t) = 5cos^3(\pi t)\)
HINT - pie and t are constant
use the trig formulas for derivatives
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
where do i take the derivative first for this function?
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.
use the chain rule , not the product rule
ok so whta rule would i be using in this case when i take ther derivative of the outermost function
The power rule :) and then as Unkle mentioned, the chain rule after that.
ok..... so using the power rule in order to find the outermost function first this would be 15?
\[15(\cos(\pi t))^2 * (d/dt)(\cos(\pi t))\] 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.
\[(f\circ h)'(x)=f(h(x))'=f'(h(x))\cdot h'(x)\]
ok i got this step now i must find the derivative of (cos(piet))^2 right?
Don't look at the square anymore. You already dealt with that outermost function. the inner function is cos(pi t)
ok so i find the derivative of that and that is -sin(piet) right?
\[15(\cos(\pi t))^2*(-\sin(\pi t))*(d/dt)(\pi t)\] Good! now we have one last step. there is an inner inner function that you need to differentiate still, as I have written above.
ok! thanks! ummm... would this be no wait u lost me
The innermost function is (pi t). So you need to multiply by the derivative of that :D Find the derivative! :D
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?
\[(d/dt)(\pi t)=\pi(d/dt)t=?\]
\(\pi\) is a constant, and when a constant combined to a variable, the variable goes to 1 but the constant stay.
im sorry u guys lost me completely
Let's restart everything, haha
nooo don't get lost :D you were so close!
\[g(t) = 5\cos^3(\pi t)\] \[f(x)=5\cos^3(x)\] \[h(x)= \pi x\]
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 \[3*5(cos^2(\pi t))*\frac{d}{dt}cos^2(\pi t)*\frac{d}{dt}(\pi t)\]
\(\large 3*5(cos^2(\pi t))*\frac{d}{dt}cos(\pi t)*\frac{d}{dt}(\pi t)\)*** Sorry
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
ok !!!! i think i got!!! THANKS SOOOOO MUCH I WAS HAVING TROUBLEE!!!!!!