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Mathematics 23 Online
OpenStudy (anonymous):

Find an equation of teh line tangent to the curve y=e^x *cosx at (0,1)

OpenStudy (anonymous):

take the derivative, replace x by 0 to get the slope, and then use slope intercept form of the line

OpenStudy (anonymous):

im confused when i take teh derivative what to do....i know f' = e^x and g' = -sinx after putting it into the product rule i get e^x (cosx) + e^x (-sinx).....do i replace it by 0 now?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

and since \[\sin(0)=0, \cos(0)=1, e^0=1\] you get 1

OpenStudy (anonymous):

ok...so then i use teh slop intercept form with the point (0,1) adn slope m=1 for the equation: y-1 = 1(x-0)...is taht teh equation of the line tangent to teh curve

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