Differentiation? (Corrected)
d ( cos ( x ) ) d cos ( 2 0 1 5 x ) ∣ ∣ ∣ ∣ ∣ x = 2 π = ?
The answer is 4060225.
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1 solution
Moderator note:
Correct.
Bonus question : What would the answer be if I replace all the cosine functions with sine functions instead?
But why should we take a limit, when the derivative is indeterminate ? Why can't we say that the derivative of the function is undefined at that point( x = 2 π ) ? @Brian Charlesworth
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I thought the derivative was indeterminate at 2 π as well, which was why I suggested to Julian to rephrase the problem as a limit, which he did at first. However, Pi Han Goh stated below that the "Chebyshev polynomial is continuous and differentiable everywhere", so the limit was then dropped. I haven't confirmed Pi Han Goh's statement for myself yet, but I have no reason to doubt him.
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It's a very simple idea. All these standard formulas ( like d x d ( x n ) = n ⋅ x n − 1 ) or rules, like, product rule, quotient rule and chain rule are results that follows from the fundamental definition of differentiation.
Whenever we differentiate a function using these rules or formula(s), we assume that the function is differentiable throughout its entire domain (or at least at the point where the value of its derivate is to be determined ).
Now just because the value of the derivate at a point comes out to be undefined then this doesn't mean at all that the function is not differentiable at that point, if that would have been the case then using standard formula(s) and rules is wrong at the first place as they can be used only when the function under consideration is differentiable.
We've to check whether LHD and RHD of f ( x ) are equal or not at x = 2 π (that's what you did). If they're same then f ( x ) is differentiable at x = 2 π and it's value can be determined.
@Julian Poon Nice problem, but I do have a concern about evaluating this expression at x = 2 π . It does go to the posted answer as x → 2 π , so you may need to consider phrasing the problem as a limit.
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Yeah ok. Btw, this problem also shows that P n ′ ( 1 ) = n 2 , where P n ( x ) is the n th chebychev polynomial of first kind.
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Oh, I didn't know that. I should spend some time reviewing chebyshev polynomials, then. :)
You don't actually need to put the limit, the chebyshev polynomial is continuous and differentiable everywhere.
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@Pi Han Goh – Well... I was just not sure...
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@Julian Poon – You can express cos ( 2 0 1 5 x ) as a sum of powers of cos ( x ) , which is differentiable by cos ( x ) at x = 2 π , right?
How about using L'Hopitâls rule to a differentiable result?
2015 is the answer
I did the same way
By the chain rule we have that d u d ( f ( x ) ) = d x d u d x d f , where u = u ( x ) .
In this case f ( x ) = cos ( 2 0 1 5 x ) and u ( x ) = cos ( x ) , so
d ( cos ( x ) ) d cos ( 2 0 1 5 x ) = − sin ( x ) − 2 0 1 5 sin ( 2 0 1 5 x ) = 2 0 1 5 ∗ sin ( x ) sin ( 2 0 1 5 x ) .
Now this cannot be evaluates at x = 2 π since it will be in the indeterminate form 0 0 , but we can evaluate it in the limit as x → 2 π . Letting u = x − 2 π , we have that
lim x → 2 π sin ( x ) sin ( 2 0 1 5 x ) = lim u → 0 sin ( u + 2 π ) sin ( 2 0 1 5 ( u + 2 π ) ) =
lim u → 0 sin ( u ) sin ( 2 0 1 5 u ) = 2 0 1 5 ∗ lim u → 0 2 0 1 5 u sin ( 2 0 1 5 u ) ∗ lim u → 0 sin ( u ) u = 2 0 1 5 ,
since lim θ → 0 θ sin ( θ ) = 1 . Thus the given expression, in the limit as x → 2 π , goes to 2 0 1 5 2 = 4 0 6 0 2 2 5 .