# tangent line or something.

• Oct 12th 2012, 08:42 AM
kaw95
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
• Oct 12th 2012, 08:47 AM
MaxJasper
Re: tangent line or something.
Tangent: y - y(0)=y'(0)*(x-x0)

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

y'(x) = cos(x) - (1+x)*sin(x)
• Oct 12th 2012, 08:50 AM
TheEmptySet
Re: tangent line or something.
Quote:

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

$\displaystyle y'(x)=-(1+x)\sin(x)+\cos(x)$

Now we can find the slope at the point (0,1)

$\displaystyle m=y'(0)=-(1+0)\sin(0)+\cos(0)=1$

Since we know the y intercept (the point (0,1) )

We know b=1

This gives

$\displaystyle y=1(x)+1 \implies y=x+1$

P.S where in Oregon are you from? I do miss Corvallis and Portland quite alot! Go Beavers!!
• Oct 12th 2012, 08:53 AM
johnsomeone
Re: tangent line or something.
"the derivative is the slope of the tangent line"
That's perhaps the single most important and commonly used phrase in Calculus I.