Find f'(x) of f(x) = 1/sqrt(x-3)

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To find f'(x) for f(x) = 1/sqrt(x-3) using the formal definition, the limit is set up as f'(x) = lim ((1/sqrt(x + h - 3)) - (1/sqrt(x - 3))) / h. The next step involves multiplying the numerator and denominator by the greatest common factor, sqrt(x-3)*sqrt(x-3+h), followed by using the conjugate of the new numerator, sqrt(x-3) + sqrt(x-3+h). This process eliminates the radicals in the numerator and cancels the h terms. Ultimately, the derivative is found to be f'(x) = -1/[2*(x-3)^(3/2)].
JohnnyPhysics
I need to find f'(x) of f(x) = 1/sqrt(x-3) using the formal definition. I set the equation up as:
f'(x) = lim ((1/ sqrt(x + h -3)) - (1/sqrt(x-3)))/h and I am not sure what the next step is...
 
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To solve the limit, multiply the top and bottom of the fraction by the GCF [sqrt(x-3)*sqrt(x-3+h)]. Then multiply the top and bottom of the fraction by the conjugate of the new numerator
[sqrt(x-3)+ sqrt(x-3+h)]. The radicals on the numerator will disappear, and the h's will cancel. Then you can substitute h=0 to find the limit. The answer is f'(x)= -1/[2*(x-3)^(3/2)].
 

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