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Calculus and Beyond Homework Help
Find equation tangent to y=(x^2)lnx at (1,0)
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[QUOTE="rmiller70015, post: 4642038, member: 281740"] [h2]Homework Statement [/h2] Find an equation for the line y=(x^2)lnx at (1,0) [h2]Homework Equations[/h2] [h2]The Attempt at a Solution[/h2] I took the first derivative of the equation and got y'=2xlnx +x In other equations, it would be simple to find the slope, but at this point I am lost, is the slope 2? It doesn't appear to be in the form of y=mx+b. So I've decided to use f'(1)=2(1)ln1 + 1 = 1 , this gives me the slope of the function at the point (1,f(1)) right? If that's the case, then y=m(x-x_1) + y_1 = 1(x-1) + 0 y=x-1 Just wanted to know if I am close or way off or maybe correct. [/QUOTE]
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Calculus and Beyond Homework Help
Find equation tangent to y=(x^2)lnx at (1,0)
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