How do you differentiate #y = cos(cos(cos(x)))#?

Answer 1

#dy/dx = -sin(cos(cos(x)))sin(cos(x))sin(x)#

This is an initially daunting-looking problem, but in reality, with an understanding of the chain rule, it is quite simple.

We know that for a function of a function like #f(g(x))#, the chain rule tells us that: #d/dy f(g(x)) = f'(g(x)g'(x)#
By applying this rule three times, we can actually determine a general rule for any function like this one where #f(g(h(x)))#: #d/dy f(g(h(x))) = f'(g(h(x)))g'(h(x))h'(x)#
So applying this rule, given that: #f(x) = g(x) = h(x) = cos(x)# thus #f'(x) = g(x) = h(x) = -sin(x)#
yields the answer: #dy/dx = -sin(cos(cos(x)))sin(cos(x))sin(x)#
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Answer 2

To differentiate ( y = \cos(\cos(\cos(x))) ), you can use the chain rule. The derivative is given by:

[ \frac{dy}{dx} = -\sin(x) \cdot \sin(\cos(x)) \cdot \sin(\cos(\cos(x))) \cdot \cos(\cos(\cos(x))) ]

This result is obtained by applying the chain rule repeatedly, from the outside function to the inside functions.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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