If #f(x) =-sqrt(2x-1) # and #g(x) = (2-1/x)^2 #, what is #f'(g(x)) #?
Depending on your teacher, this is likely as far as you'll need to go.
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To find ( f'(g(x)) ), you need to find the derivative of ( f ) with respect to ( x ) and then substitute ( g(x) ) in place of ( x ).
First, find ( f'(x) ): [ f'(x) = \frac{d}{dx}(-\sqrt{2x-1}) ] [ f'(x) = -\frac{1}{2\sqrt{2x-1}} ]
Now, substitute ( g(x) ) into ( f'(x) ): [ f'(g(x)) = -\frac{1}{2\sqrt{2g(x)-1}} ]
Substitute ( g(x) = (2 - \frac{1}{x})^2 ): [ f'(g(x)) = -\frac{1}{2\sqrt{2(2 - \frac{1}{x})^2 - 1}} ]
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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.
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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