## v.s 3 years ago Find equations of the tangent line and normal line to the curve at the given point. y = x^4 + 4e^x, (0, 4) tangent line y = normal line y =

1. campbell_st

can you find the derivative of the curve...?

2. v.s

i don't know

3. campbell_st

ok... have you done or are you doing calculus?

4. v.s

doing calculus

5. campbell_st

ok... can you differentiate y = x^4...?

6. v.s

4x^3

7. campbell_st

and can you differentiate \[ y = 4e^{x} \]

8. v.s

x4e

9. v.s

or is it x4e^-x

10. campbell_st

quick lesson \[y = 4e^{f(x)} ..... then .... y'=4f'(x) e^{f(x)}\] and in your question f(x) = x... try the derivative again for \[y = 4e^x\]

11. v.s

4xe^-x

12. campbell_st

not quite if f(x) = x f'(x) = 1 so its y' = 4e^x so you you function the derivative is \[y' = 4x^3 + 4e^x\] ok... you now have the equation for the slop of the tangent...(1st derivative) subsitiute x = 0 to find the slope of the tangent at the point (0, 4)

13. v.s

y'=4

14. campbell_st

thats correct so you have a tangent or line with a slope of 4 and point (0, 4) can you find the equation of the line...? for the normal... the slope will be the negative reciprocal of the slope of the tangent... what would the slope of the normal be..?

15. v.s

=-1/4x^3+4e^x

16. campbell_st

no... ok... lets step back... you have a slope of 4 and point (0, 4) the equation of the tangent will be y = 4x + b or using the formula \[y - 4 = 4(x - 0)\] whats the equation of the tangent...?

17. v.s

y=4x-4

18. campbell_st

can you try again..

19. v.s

y=4x+4

20. campbell_st

ok... thats the equation of the tangent... now the normal... which is perpendicular to the tangent... the slope is the negative reciprocal of the tangent... so if the tangent has a slope of 4 ... what is the slope of the normal..?

21. v.s

y=-1/4x+4

22. campbell_st

that looks good... so there you have your solutions.

23. v.s

thank you

24. campbell_st

hope it helps... you do need to practice you differentiation basics.

25. v.s

yeaa