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\[y=e ^{2x}\cos \pi x\](0,1)

why do you differentiate again?

you pretty much instantly get \(m=2\)

\((fg)'=f'g+g'f\)
you have a product, you need the product rule

yeah you need the slope and a point

slope is the value of the derivative evaluated at the \(x\) value , in your case 0

ok so once i have the slope do i plug it into
\[y-y _{1}=m(x-x _{1})\]

with the (0,1)

so the answer should be
\[y=2x+1\]
right? = )

yes, if the value of the derivative at (0,1) is 2

woo-hoo! I just needed a friendly reminder lol = )