## Zenmo one year ago Find an equation of the tangent line to the curve at the given point (Calculus Derivatives).

1. Zenmo

Question #1

2. freckles

ok what did you find for y'?

3. Zenmo

$\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h }$ = $\lim_{h \rightarrow 0}\frac{ [(3+h)^3-3(3+h)+1-(3^3-3(3)+1)] }{ h }$ . Putting more work on here second

4. FireKat97

do you have to apply differentiation by first principles? :/

5. Zenmo

$(3+h)^3; (3+h)(3+h)=9+3h+3h+h^2=9+6h+h^2 ; (9+6h+h^2)(3+h) = 27+18h+3h^2+9h+6h^2+h^3$

6. anonymous

Are you familiar with the power rule of differentiation? $\frac{ d }{ dx }[x^n]=n*x^{n-1}$

7. freckles

you might like pascal's triangle 4th row says 1 3 3 1 $(3+h)^3=1(3)^3+3(3)^2h+3(3)h^2+h^3$

8. Zenmo

Our class isn't on there yet, this is the beginning of calculus for the end of Chapter 2

9. freckles

$\lim_{h \rightarrow 0} \frac{3^3+3(3)^2h+3(3)h^2+h^3-3(3+h)+1-(3^3-3(3)+1)}{h}$

10. Zenmo

Sec. I might found the answer.

11. Zenmo

I gotten 6x+1 as the answer.

12. freckles

ok let's go slow then you do see above the 3^3 's cancel right?

13. freckles

since 3^3-3^3=0 and also you should see the 3(3) also known as 9's cancel you know since -9+9=0

14. Zenmo

This is how the book does it. Similar problem.

15. freckles

you should also see the 1's cancel since 1-1=0

16. freckles

$\lim_{h \rightarrow 0} \frac{3^3+3(3)^2h+3(3)h^2+h^3-3(3+h)+1-(3^3-3(3)+1)}{h} \\ \lim_{h \rightarrow 0} \frac{3(3)^2h+3(3)h^2+h^3-3h}{h}$

17. freckles

now divide top and bottom by h

18. Zenmo

$\lim_{h \rightarrow 0}(h^2+9h+27)$ Did u get that?

19. freckles

should wind up with $\lim_{h \rightarrow 0} (3(3)^2+3(3)h+h^2-3)$

20. freckles

something happened to your -3 part above

21. freckles

should be $\lim_{h \rightarrow 0} (h^2+9h+27-3)$

22. freckles

so your slope on your homework is off which will also make your y-intercept you found off too

23. freckles

please let me know if I need to redo the algebra for you if you don't understand

24. freckles

I guess you can call it algebra/arithmetic whatever that stuff is

25. Zenmo

$\lim_{h \rightarrow 0}(h^2+9h-24)$ that correct? I'm looking at calculations at the moment.

26. freckles

well 27-3 is 24 not -24 :p

27. Zenmo

By using the point (3,19) and using the point-slope form for slope M=24. I get $y=24x+19$

28. freckles

y-19=f'(3)(x-3) y-19=24(x-3) y-19=24x-72 hmmm...-72+19...

29. Zenmo

$y=24x+53$

30. freckles

I hate to be that guy but -72+19 is -53

31. Zenmo

Yea so the answer is $y=24x-53$ ?

32. freckles

yep

33. freckles

do you understand how we got the slope to be 24?

34. Zenmo

Yea, turns out I was messing up on simple arithmetic/algebra breaks for: adding/subtracting/factoring etc.

35. Zenmo

Yea I understood the core concepts, just messed up with the things from above.

36. Zenmo

Time for a study break haha

37. Zenmo

Thanks a lot for sticking around and helping. :)

38. freckles

we all mess on arithmetic just go pet a cat and then come back you will be already (that is my only study advice; cats for the win!)