If #f(x) =-sqrt(2x-1) # and #g(x) = 3/x^3 #, what is #f'(g(x)) #?

Answer 1

Cameron Alexander

Not the chain rule

that's not the chain rule

The chain rule is:

#(f\circ g)'=(f'\circ g)\cdot g'#

#(f\circ g)' ne f'(g)#

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Answer 2

Charles Anctil

To find ( f'(g(x)) ), we first need to find the derivative of ( f(x) ) with respect to ( x ) and then substitute ( g(x) ) into that derivative expression.

Given ( f(x) = -\sqrt{2x - 1} ) and ( g(x) = \frac{3}{x^3} ),

[ f'(x) = \frac{d}{dx}(-\sqrt{2x - 1}) = -\frac{1}{2\sqrt{2x - 1}} ]

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

[ f'(g(x)) = -\frac{1}{2\sqrt{2g(x) - 1}} = -\frac{1}{2\sqrt{2(\frac{3}{x^3}) - 1}} = -\frac{1}{2\sqrt{\frac{6}{x^3} - 1}} ]

So, ( f'(g(x)) = -\frac{1}{2\sqrt{\frac{6}{x^3} - 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.