## bettyboop8904 Group Title find an equation of the tangent line to curve at the given point: y=e^(2x)cos(pi)x (0,1) one year ago one year ago

1. bettyboop8904 Group Title

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

2. satellite73 Group Title

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

3. bettyboop8904 Group Title

why do you differentiate again?

4. satellite73 Group Title

you pretty much instantly get $$m=2$$

5. satellite73 Group Title

?

6. satellite73 Group Title

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

7. satellite73 Group Title

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

8. bettyboop8904 Group Title

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?

9. satellite73 Group Title

yeah you need the slope and a point

10. satellite73 Group Title

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

11. bettyboop8904 Group Title

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

12. bettyboop8904 Group Title

@satellite73

13. RyanL. Group Title

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.

14. sweet1137 Group Title

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

15. bettyboop8904 Group Title

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

16. bettyboop8904 Group Title

with the (0,1)

17. bettyboop8904 Group Title

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

18. sweet1137 Group Title

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

19. bettyboop8904 Group Title

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