Mathematics OpenStudy (anonymous):

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) OpenStudy (anonymous):

To find the slope of the tangent line at a point, find the value of the derivative at that point. OpenStudy (anonymous):

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

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

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. OpenStudy (anonymous):

Are you still using the limit definition of the derivative? OpenStudy (anonymous):

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... OpenStudy (anonymous):

Sorry, hadn't realized. Give me a minute, I'll type something up. OpenStudy (anonymous):

that's what they want me to do. Use the limiting process to find the slope. OpenStudy (anonymous):

$\large\lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\lim_{h\to0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}=\lim_{h\to0}\frac{\frac{-h}{x(x+h)}}{h}=\lim_{h\to0}\frac{-1}{x(x+h)}$Can you take it from there? OpenStudy (anonymous):

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

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? OpenStudy (anonymous):

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

$\large\lim_{h\to0}\frac{\frac{1}{1+h}-\frac{1}{1}}{h}=\lim_{h\to0}\frac{\frac{1-(1+h)}{1(1+h)}}{h}=\lim_{h\to0}\frac{\frac{-h}{1(1+h)}}{h}=\lim_{h\to0}\frac{-1}{1(1+h)}$ OpenStudy (anonymous):

Okay. thanks! OpenStudy (anonymous):

:)

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