What is the second derivative of #f(cosx)# when #x=pi/2# where #f(x)=sinx#?

Answer 1

#0#

We have: #f(x) = sin(x)#
#Rightarrow f(cos(x)) = sin(cos(x))#

Let's differentiate this function:

#Rightarrow f'(cos(x)) = frac(d)(dx)(sin(cos(x)))#

We can differentiate this using the chain rule.

Let #u = cos(x) Rightarrow u' = - sin(x)# and #v = sin(u) Rightarrow v' = cos(u)#:
#Rightarrow f'(cos(x)) = u' cdot v'#
#Rightarrow f'(cos(x)) = - sin(x) cdot cos(u)#
#Rightarrow f'(cos(x)) = - sin(x) cos(u)#
Replace #u# with #cos(x)#:
#Rightarrow f'(cos(x)) = - sin(x) cos(cos(x))#

We need to find the second derivative, so let's differentiate this one more time:

#Rightarrow f''(cos(x)) = frac(d)(dx)(- sin(x) cos(cos(x)))#
#Rightarrow f''(cos(x)) = - frac(d)(dx)(sin(x) cos(cos(x)))#

We can differentiate this using a combination of the product rule and the chain rule:

#Rightarrow f''(cos(x)) = sin(x) cdot frac(d)(dx)(cos(cos(x))) + cos(cos(x)) cdot frac(d)(dx)(sin(x))#
Let #u = cos(x) Rightarrow u' = - sin(x)# and #v = cos(u) Rightarrow v' = - sin(u)#:
#Rightarrow f''(cos(x)) = sin(x) cdot u' cdot v' + cos(cos(x)) cdot cos(x)#
#Rightarrow f''(cos(x)) = sin(x) cdot (- sin(x)) cdot (- sin(u)) + cos(x) cos(cos(x))#
#Rightarrow f''(cos(x)) = sin^(2)(x) cdot sin(u) + cos(x) cos(cos(x))#
Replace #u# with #cos(x)#:
#Rightarrow f''(cos(x)) = sin^(2)(x) cdot sin(cos(x)) + cos(x) cos(cos(x))#
#therefore f''(cos(x)) = sin^(2)(x) sin(cos(x)) + cos(x) cos(cos(x))#
Finally, let's evaluate #f''(cos(x))# at #x = frac(pi)(2)#:
#Rightarrow f''(cos(frac(pi)(2))) = sin^(2)(frac(pi)(2)) sin(cos(frac(pi)(2))) + cos(frac(pi)(2)) cos(cos(frac(pi)(2)))#
#Rightarrow f''(cos(frac(pi)(2))) = 1 cdot sin(0) + 0 cdot cos(0)#
#Rightarrow f''(cos(frac(pi)(2))) = 1 cdot 0 + 0 cdot 1#
#Rightarrow f''(cos(frac(pi)(2))) = 0 + 0#
#therefore f''(cos(frac(pi)(2))) = 0#
Therefore, the final answer is #0#.
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Answer 2

# [f''(cos(x))]_(x=pi/2) = 0 #

We have:

# f(x)=sinx #
And we want the second derivative of #f(cosx)#

Let:

# g(x) = f(cosx) = sin(cosx) #

Then by the chain rule we have:

# g'(x) = d/dx sin(cosx) # # " " = (cos(cosx))(-sinx) # # " " = -sinx \ cos(cosx) #

And for the second derivative we also need the product rule'

# g''(x) = d/dx -sinx \ cos(cosx) # # " " = - \ d/dx sinx \ cos(cosx) # # " " = - { (sinx)(d/dx cos(cosx)) + (d/dx sinx)(cos(cosx) }#
# " " = - { (sinx)(-sinx(cosx)(-sinx) + (cosx)(cos(cosx) }#
# " " = - sin^2x \ sinx(cosx) - cosx \ cos(cosx) #
And with #x=pi/2# we have:
# g(pi/2) = - sin^2(pi/2) \ sinx(cos(pi/2)) - cos(pi/2) \ cos(cos(pi/2)) #
And as #sin(pi/2)=1#, #cos(pi/2)=0# and #sin0=0# we have
# g(pi/2) = - (1) \ sin(0) - (0) \ cos(0) # # " " = - (1) \ (0) # # " " = 0 #
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Answer 3

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.

  1. 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 ]

  2. 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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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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