What is the second derivative of #f(cosx)# when #x=pi/2# where #f(x)=sinx#?
Let's differentiate this function:
We can differentiate this using the chain rule.
We need to find the second derivative, so let's differentiate this one more time:
We can differentiate this using a combination of the product rule and the chain rule:
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We have:
Let:
Then by the chain rule we have:
And for the second derivative we also need the product rule'
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To find the second derivative of ( f(\cos x) ) when ( x = \frac{\pi}{2} ) where ( f(x) = \sin x ), we first find the first derivative of ( f(\cos x) ) using the chain rule, then take its derivative again to find the second derivative.

First derivative of ( f(\cos x) ): [ f'(\cos x) = f'(u) \cdot u' ] [ f'(u) = \cos x ] [ u' = \sin x ] [ f'(\cos x) = \cos x \cdot (\sin x) = \cos x \sin x ]

Second derivative of ( f(\cos x) ): [ f''(\cos x) = \frac{d}{dx}(\cos x \sin x) ] Using the product rule: [ f''(\cos x) = (\sin^2 x  \cos^2 x) ] Now, when ( x = \frac{\pi}{2} ): [ \sin^2 \left(\frac{\pi}{2}\right)  \cos^2 \left(\frac{\pi}{2}\right) = 1  0 = 1 ] So, the second derivative of ( f(\cos x) ) when ( x = \frac{\pi}{2} ) is ( \boxed{1} ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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