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

[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...

 

[StephenPrivitera]

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)].

 

[tacman]

To find the derivative of f(x) using the formal definition, we need to use the limit definition of derivative:

f'(x) = lim(h->0) (f(x+h) - f(x))/h

In this case, f(x) = 1/sqrt(x-3). So we have:

f'(x) = lim(h->0) ((1/sqrt(x+h-3)) - (1/sqrt(x-3)))/h

Next, we need to simplify the expression inside the limit. To do this, we can use the difference of squares formula to simplify the numerator:

f'(x) = lim(h->0) ((1/sqrt(x+h-3)) - (1/sqrt(x-3))) * (sqrt(x+h-3) + sqrt(x-3))/h

Now, we can use the conjugate of the first term in the numerator to simplify further:

f'(x) = lim(h->0) (((1/sqrt(x+h-3)) - (1/sqrt(x-3))) * (sqrt(x+h-3) + sqrt(x-3)) * (sqrt(x+h-3) - sqrt(x-3)))/h * (sqrt(x+h-3) - sqrt(x-3))

Simplifying this, we get:

f'(x) = lim(h->0) (1 - sqrt(x+h-3) * sqrt(x-3))/h * (sqrt(x+h-3) - sqrt(x-3))

Now, we can apply the limit as h->0:

f'(x) = (1 - sqrt(x-3)^2)/ (sqrt(x-3) * (sqrt(x-3) - sqrt(x-3)))

Simplifying this, we get:

f'(x) = -1/ (sqrt(x-3)^3)

So, the derivative of f(x) = 1/sqrt(x-3) using the formal definition is f'(x) = -1/ (sqrt(x-3)^3).