Differentiation

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whiteti

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Jun 3, 2013
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Evaluate f^50(7pi/12) Where f(x)=cos(2x)

I've barely managed to learn how to get the first derivative. All ive got written down is that 50 is the same as 48 +2 = 4*12+2
Help?
 
whiteti said:
Evaluate f^50(7pi/12) Where f(x)=cos(2x)

I've barely managed to learn how to get the first derivative. All ive got written down is that 50 is the same as 48 +2 = 4*12+2
Help?
Click to expand...
Do you not understand why you have "written down that 50 is the same as 48+ 2= 4*12+ 2"?

The first derivative is -2sin(2x). The second derivative is \(\displaystyle -2^2 cos(2x)\). The third derivative is \(\displaystyle 2^3 sin(2x)\). And the fourth derivative is \(\displaystyle 2^4 cos(2x)\). Now look at that as two sepate parts. The powers of 2 are, for first derivative, \(\displaystyle 2^1\), for second derivative, \(\displaystyle 2^2\), for third derivative \(\displaystyle 2^3\), etc. What do you think the 50th derivative will give you?

And the trig part: cos(2x), -sin(2x), -cos(2x), sin(2x), cos(2x) so that at the fourth derivative you are back to cos(2x). Since 48+ 2= 4*12+ 2, differentiating 50 times you will go through that cycle 12 times and the differentiate two more times. What was the second derivative?
 
Last edited:
Subhotosh Khan said:
First evaluate \(\displaystyle \theta \ = \ 2\frac{7\pi}{2}\)

Then evaluate \(\displaystyle f(\frac{7\pi}{2}) \ = \ cos(\theta)\)


Then evaluate \(\displaystyle \left [f(\frac{7\pi}{2})\right ]^{50} \ \)
Click to expand...
That is not my understanding of the problem. From what the OP says, " I've barely managed to learn how to get the first derivative. All ive got written down is that 50 is the same as 48 +2 = 4*12+2" I think he means the 50th derivative, not the 50th power.
 
whiteti said:
Evaluate f^50(7pi/12) Where f(x)=cos(2x)

I've barely managed to learn how to get the first derivative. All ive got written down is that 50 is the same as 48 +2 = 4*12+2
Help?
Click to expand...
I interpret - especially looking at your start at doing it - that the ^50 means 50th derivative.

a) every time you differentiate, you will multiply by 2 from the 2x. Thus one factor in the result will be 2^50.

b) anytime you differentiate a sine or cosine 4 times, you are back where you started. That is why you can omit 48 of the differentiations! What do you have if you differentiate a cosine twice?

c) after differentiating, evaluate at x=(7 p1)/12, or 2x=(7 pi)/6

Let us see how far this gets you...
 
so I should have f^50=-2^50cos(2x)?

where does the 7pi/12 come in?
 
You have a formula for \(\displaystyle f^{50}(x)\) so to find \(\displaystyle f^{50}(7\pi/12)\), replace x with \(\displaystyle 7\pi/12\).
 
whiteti said:
so my answer should be

f^(50)(7pi/12) = 2^50 cos(2(7pi/12))
?
Click to expand...
You had a minus sign in your formula - don't leave it out.

You can simplify 2(7pi/12) = 7pi/6,

And you should be able to determine the exact value of the cosine. Where is 7pi/6 on the unit circle?
 
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