Thread: tangent line or something.

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

Tangent: y - y(0)=y'(0)*(x-x0)

{x0,y(x0)} = {0,1}

y'(x) = cos(x) - (1+x)*sin(x)

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.