If #f(x)= sqrt(x^21 # and #g(x) = 1/x #, what is #f'(g(x)) #?
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To find ( f'(g(x)) ), you first need to find the derivative of ( f(x) ) with respect to ( x ), and then substitute ( g(x) ) for ( x ) in the resulting expression.

Derivative of ( f(x) = \sqrt{x^2  1} ) with respect to ( x ): [ f'(x) = \frac{1}{2\sqrt{x^2  1}} \cdot 2x = \frac{x}{\sqrt{x^2  1}} ]

Substitute ( g(x) = \frac{1}{x} ) into ( f'(x) ): [ f'(g(x)) = \frac{g(x)}{\sqrt{g(x)^2  1}} = \frac{\frac{1}{x}}{\sqrt{\left(\frac{1}{x}\right)^2  1}} ]
[ f'(g(x)) = \frac{\frac{1}{x}}{\sqrt{\frac{1}{x^2}  1}} = \frac{\frac{1}{x}}{\sqrt{\frac{1  x^2}{x^2}}} ]
[ f'(g(x)) = \frac{\frac{1}{x}}{\frac{\sqrt{1  x^2}}{x}} = \frac{1}{x} \cdot \frac{x}{\sqrt{1  x^2}} ]
[ \therefore f'(g(x)) = \frac{x}{x\sqrt{1  x^2}} ]
Note: Since ( g(x) = \frac{1}{x} ), ( x ) cannot be zero, hence ( x \neq 0 ). Thus, the absolute value ( x ) is redundant. Therefore, the final expression can be simplified to:
[ \boxed{f'(g(x)) = \frac{1}{x\sqrt{1  x^2}}} ]
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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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