Can anyone help me work through this precalc problem? Find the slope of the tangent line to the graph of f(x) = (1/x) at the point (1,1)

To find the slope of the tangent line at a point, find the value of the derivative at that point.

I get mixed up in the simplification. I've been able to do the others, but the fraction ones throw me everytime.

Rewrite the function as \(f(x)=x^{-1}\) if that makes it easier.

I don't think I am doing this right. I get to (1/1+h) -1/h and I get stuck. I think that's where I go wrong each time.

Are you still using the limit definition of the derivative?

how is this a precalc problem? you'll need to take the derivative... unless you plan on finding the slope of the tangent line at x=1 using the limit process...

Sorry, hadn't realized. Give me a minute, I'll type something up.

that's what they want me to do. Use the limiting process to find the slope.

\[ \large\lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\lim_{h\to0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}=\lim_{h\to0}\frac{\frac{-h}{x(x+h)}}{h}=\lim_{h\to0}\frac{-1}{x(x+h)} \]Can you take it from there?

I think so. Any tips about how to convert this to the limiting process my text wants us to use?

I don't know what your text wants you to use, sorry. This is a limiting process. Does your text want you to graph secant lines or something?

Sorry. It wants me to plug in the values for x at the given point. So (1/x+h) becomes (1/1+h)

\[\large\lim_{h\to0}\frac{\frac{1}{1+h}-\frac{1}{1}}{h}=\lim_{h\to0}\frac{\frac{1-(1+h)}{1(1+h)}}{h}=\lim_{h\to0}\frac{\frac{-h}{1(1+h)}}{h}=\lim_{h\to0}\frac{-1}{1(1+h)}\]

Okay. thanks!

:)

Join our real-time social learning platform and learn together with your friends!