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 

Re: tangent line or something.
Tangent: y  y(0)=y'(0)*(xx0)
{x0,y(x0)} = {0,1}
y'(x) = cos(x)  (1+x)*sin(x)
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!!
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.