find an equation of the tangent line to curve at the given point:

y=e^(2x)cos(pi)x (0,1)

Question

find an equation of the tangent line to curve at the given point:

y=e^(2x)cos(pi)x          (0,1)

anonymous · Answer

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

anonymous · Answer

$$y'=2e^{2x}\cos(\pi x)-\pi e^{2x}\sin(\pi x)$$ by the product and chain rule
replace $x$ by $0$ and find your slope

anonymous · Answer

why do you differentiate again?

anonymous · Answer

you pretty much instantly get $m=2$

anonymous · Answer

?

anonymous · Answer

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

anonymous · Answer

$f(x)=e^{2x},f'(x)=2e^{2x}$ by the chain rule
$$g(x)=\cos(\pi x), g'(x)=-\pi \sin(\pi x)$$ again by the chain rule

anonymous · Answer

no no i understand that but you're looking for the slope so what connection does the equation and the derivative have? The slope is connected nvm i answered my own question lol is that what you do every time you want to find an equation of a tangent line like this? is you take the derivative?

anonymous · Answer

yeah you need the slope and a point

anonymous · Answer

slope is the value of the derivative evaluated at the $x$ value , in your case 0

anonymous · Answer

ok so from the beginning again... you said you take the derivative of the function and then plug 0 in for x?

anonymous · Answer

@satellite73

anonymous · Answer

Basically the function is curved and at the given point we want to know what it looks like if we zoom in on the graph. As we zoom in to that point the graph starts to look like a straight line with a slope. In order to find that slope we find the derivative of the function and plug the x value of the point we zoom in. The output is out slope. Once we have that we can use point slope formula to find the equation.

anonymous · Answer

the tangent line to a curve at a point represents the rate of change of that function at that point. The tangent line not only passes through that point but extends outward with a slope that is equal to the rate of change of the function at that point

anonymous · Answer

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

anonymous · Answer

with the (0,1)

anonymous · Answer

so the answer should be
$$y=2x+1$$
right? = )

anonymous · Answer

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

anonymous · Answer

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