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Mathematics 15 Online
OpenStudy (anonymous):

@zepdrix Find the derivative of the function. f(t)=cos^6(pi t -14)

zepdrix (zepdrix):

\[\large f(t)=\cos^6\left(\pi t-14\right)\] Are you able to recognize what the outermost function is? The notation we use for trig functions can make that a little difficult sometimes. Writing it this way might help if you're having any trouble with that particular idea.\[\large f(t)=\left[\cos\left(\pi t-14\right)\right]^6\]

zepdrix (zepdrix):

So to take the derivative, we will apply the power rule to the `outermost function`. Then we'll multiply by the derivative of the inner function.

zepdrix (zepdrix):

Power rule gives us this, right? \[\large f'(t)=6\cos^5\left(\pi t-14\right) \color{royalblue}{\left[\cos(\pi t-14)\right]'}\] Chain rule tells us to multiply by that blue term. We still need to take the derivative of that blue term. And confusion yet? :O

OpenStudy (anonymous):

So far so good!

zepdrix (zepdrix):

\[\large f'(t)=6\cos^5\left(\pi t-14\right) \color{orangered}{\left[-\sin(\pi t-14)\right]}\color{royalblue}{\left(\pi t-14\right)'}\] Cosine gives us -sine right? We then apply the chain rule once more, multiplying by the derivative of the inner inner function.

OpenStudy (anonymous):

Mmkay

OpenStudy (anonymous):

Why do we keep adding on (pi t -14)?

zepdrix (zepdrix):

Ok from this step, \[\large f'(t)=6\cos^5\left(\pi t-14\right) \color{royalblue}{\left[\cos(\pi t-14)\right]'}\] The pi*t-14 is inside of the trig function. When we apply the chain rule, mmm trying to think of the right way to explain this... We're making a copy of the cosine function (which includes it's innards), and taking its derivative.

zepdrix (zepdrix):

Confused by the next (pi*t-14) that shows up after this one?

OpenStudy (anonymous):

I think I got it

zepdrix (zepdrix):

The derivative of that last portion gives us, \[\large f'(t)=6\cos^5\left(\pi t-14\right)\left[-\sin(\pi t-14)\right]\color{orangered}{\left(\pi\right)}\]

OpenStudy (anonymous):

okay

zepdrix (zepdrix):

Since multiplication is `commutative`, we can rearrange the order in which we multiply. This is just to make it look a little cleaner.\[\large f'(t)=-6\pi \cos^5(\pi t-14)\sin(\pi t-14)\]

zepdrix (zepdrix):

Chain rule is a little tricky to get used to. Practice more of these!! :)

OpenStudy (anonymous):

who that's the final answer! wow you are so helpful! And I know I've been having trouble with chain rule :(

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