What is #f'(-pi/3)# when you are given #f(x)=sin^7(x)#?

Answer 1
It is #(7sqrt3)/2^7=(7sqrt3)/128#

Method

#f(x)=sin^7(x)#
It is very useful to re-write this as #f(x)=(sin(x))^7# because this makes it clear that what we have is a #7^(th)# power function.

Apply the chain rule and the power rule (also known as the generalized power rule in some circles).

For #f(x)=(g(x))^n#, the derivative is #f'(x)=n(g(x))^(n-1)*g'(x)#,
In other notation #d/(dx)(u^n)=n u^(n-1) (du)/(dx)#
In either case, for your question #f'(x)=7(sin(x))^6*cos(x)#
You could write #f'(x)=7sin^6(x) *cos (x)#
At #x=- pi/3#, we have #f'(- pi/3)=7sin^6(- pi/3) *cos (- pi/3)=7(1/2)^6(sqrt3/2)=(7sqrt3)/2^7#
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Answer 2
#"let " y= f(x) # #=> dy/dx = f'(x)#
#=>y = sin^7(x)#
#"let " u = sin(x) => y = u^7#
#du/dx = cos(x)#
# dy/du = 7*u^6#
Now, #f'(x) = (dy)/(dx)# #= (dy)/(du)*(du)/(dx)# {Do you agree?} #= 7u^6*cosx# but remember #u = sin(x)# #=> f'(x) = 7sin^6(x)cos(x)#
#=> f'(-pi/3) = 7*(sin(-pi/3))^6**cos(-pi/3)#
# = 7(-sqrt(3)/2)^6**(1/2)#

It's Your Honor To Simplify

NOTE: Are you wondering why I'm engaging in so much "let stuff"?

the reason is there are more than one function in #f(x)#
** there's : #sin^7(x)# and there's #sin(x) #!!
so to find the #f'(x)# i need to find the #f'# of #sin^7(x)# AND the #f'# of #sin(x)#
that's why i need to let # y= f(x)# then let #u = sin(x)# }
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Answer 3

To find ( f'(-\frac{\pi}{3}) ) for ( f(x) = \sin^7(x) ), we differentiate ( f(x) ) with respect to ( x ) using the chain rule:

[ f'(x) = 7\sin^6(x)\cos(x) ]

Substitute ( x = -\frac{\pi}{3} ) into the derivative function:

[ f'(-\frac{\pi}{3}) = 7\sin^6(-\frac{\pi}{3})\cos(-\frac{\pi}{3}) ]

[ f'(-\frac{\pi}{3}) = 7\left(\frac{\sqrt{3}}{2}\right)^6 \left(\frac{1}{2}\right) ]

[ f'(-\frac{\pi}{3}) = 7\left(\frac{3\sqrt{3}}{8}\right) ]

[ f'(-\frac{\pi}{3}) = \frac{21\sqrt{3}}{8} ]

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Answer 4

To find ( f'(-\frac{\pi}{3}) ) when ( f(x) = \sin^7(x) ), we first need to find the derivative of ( f(x) ), denoted as ( f'(x) ). Then, we can evaluate ( f'(-\frac{\pi}{3}) ).

Using the chain rule and the power rule, the derivative of ( f(x) = \sin^7(x) ) is:

[ f'(x) = 7 \sin^6(x) \cos(x) ]

Now, substituting ( x = -\frac{\pi}{3} ) into ( f'(x) ):

[ f'(-\frac{\pi}{3}) = 7 \sin^6(-\frac{\pi}{3}) \cos(-\frac{\pi}{3}) ]

[ = 7 \left(\frac{\sqrt{3}}{2}\right)^6 \left(\frac{1}{2}\right) ]

[ = 7 \left(\frac{3}{4}\right)^3 \left(\frac{1}{2}\right) ]

[ = 7 \left(\frac{27}{64}\right) \left(\frac{1}{2}\right) ]

[ = \frac{189}{128} ]

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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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