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MIT 18.01 Single Variable Calculus (OCW) 9 Online
OpenStudy (anonymous):

Hi there. I have been working on this question far too long and now hoping for some help. I have to find the slope of the tangent line to the function y=1/√(2x-1) at the point a, using the definition of the derivative. I set used the def'n of the derivative and came up with: [(1/√(2x+2h-1))-(1/(√x+h))]/h. My derviative ended up being -x/92x-1+√(2x-1)√x) which I do not think is correct. The reason I think it is wrong is because part b of the question asks, "at which point does the tangent to the function have a slope of -1?" I plugged the value of -1 into the slope function (which was my deriva

OpenStudy (stacey):

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