How do you find out what dy/dx is? They give me this to plug in....1/x?

Question

The book tells me to use formula 12...which is dy/dx= lim as h->0
delta y/delta x=lim as h->0 f(x+h)-f(x) / h

Answer 1

Here's a hint: Try using the formula they gave you.
dy/dx=[h→0]lim (1/(x+h)-1/x)/h
dy/dx=[h→0]lim (x/(x(x+h))-(x+h)/(x(x+h))/h
dy/dx=[h→0]lim (-h/(x(x+h)))/h
dy/dx=[h→0]lim -1/(x(x+h))
dy/dx=[h→0]lim -1/(x²+hx)
dy/dx=-1/x²

Answer 2

Try setting h at 0. Then add a little and a little more. Then set h really high and work to a lower value. It basically gives you some function minus that same function over a really really small value. The idea is 0/0. That is not the answer, that is a basic trend toward an answer. There are other methods of analysis you can try. Setting h really big gives a function of a really big number divided by a really big number. Not knowing what the function is, that is tough to determine.
If you use 1/x as the function, that helps much. Good luck!