How do you find an equation of the target line to the at the given point : y=(1+x)cosx at the point 0,1. I think you use product rule and then plug it into y=mx+b but i cant get right answer
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Tangent: y - y(0)=y'(0)*(x-x0) {x0,y(x0)} = {0,1} y'(x) = cos(x) - (1+x)*sin(x)
Last edited by MaxJasper; Oct 12th 2012 at 10:25 AM.
Originally Posted by kaw95 How do you find an equation of the target line to the at the given point : y=(1+x)cosx at the point 0,1. I think you use product rule and then plug it into y=mx+b but i cant get right answer Yes you use the product rule This gives Now we can find the slope at the point (0,1) Since we know the y intercept (the point (0,1) ) We know b=1 This gives P.S where in Oregon are you from? I do miss Corvallis and Portland quite alot! Go Beavers!!
"the derivative is the slope of the tangent line" That's perhaps the single most important and commonly used phrase in Calculus I.
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