Another calc question... derivative of 1/square root x?

Question

This has to be differentiated using f(x) = [f(x+h) - f(x)]/h and taking the limit as h tends to 0.

Ahhhhhhhhhhhcalculus!

Answer 1

lijm h->0 ((x+h)^-0.5 - x^-0.5)/h = -0.5x^-1.5

Answer 2

1/SQRT(x) is the same as x^(-1/2), because a square root is the same as "to the power of 1/2" and an inverse is the same as a negative exponent. So the answer is (-1/2)x^(-3/2).

What they're asking you to do is use the limit definition of a derivative. How to do that:
--Find f(x+h). This is simply substituting (x+h) for every x in the function.
f(x+h) (SQRT(x)) = SQRT(x+h).
--Find f(x). That's the same as the original function--SQRT(X).
--Substitute f(x+h) and f(x) into the limit definition of the derivative, like so:
lim x->0 [(SQRT(x+h))-SQRT(x)]/h
--Simplify, if necessary
--Find the limit, by replacing every x with a 0
--Simplify again. The h's should cancel out.

The result is the answer.

Don't worry--you won't be using the limit definition forever. There are simple formulas for derivatives which you can memorize, and it'll get faster after they let you start using them.

Answer 3

think of of it as x^(-one million/2) you could now use your ability rule to multiply via -one million/2 and then subtract the flexibility via one million, it extremely is going to look somewhat messy if that's what you're traumatic approximately

Answer 4

f(x)=sqrt(x^-1)
f(x+h)=1/(x+h)^.5

f(x+h)-f(x)=exapnding f(x+h) and keeping h=0
we get required value