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

Method

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

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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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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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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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