anonymous 5 years ago how do i find the derivative of: Y=x^3 tan^2 x

1. myininaya

you have a function times a function so you have to use the product rule

2. myininaya

so we have y'=[x^3]'(tanx)^2+x^3[(tanx)^2]'

3. myininaya

(x^3)'=3x^2 using power rule

4. myininaya

we have [(tanx)^2]'=2(tanx)*[secx]^2 using chain rule

5. anonymous

thank you so much

6. myininaya

np

7. anonymous

$3x^2(\tan^2x)+x^3(1/\cos^2x)$

8. anonymous

thank you

9. anonymous

anytime :)

10. myininaya

That's not right mmonish91

11. anonymous

we know the equation is uv' + u'v, so take the following: - u = x^3 - u' = 3x^2 - v = tan^2x - v' = 2 tanx sec^2x so: y' = uv' + u'v = (x^3)(2tanxsec^2x) + (3x^2)(tan^2x)

12. anonymous

^_^ simply as that.

13. anonymous

simple*

14. anonymous

15. anonymous

mmonish dear, you started correctly, but didn't get d/dx tan^2x right. the derivative of tanx = sec^2x , but in this case the derivative of: (tanx)^ 2 = 2.(tanx)(sec^2x) ; you multiply the power first then subtract it, then you derive tanx. (power rule) ^_^ a^ u = u(a)(u') <-- rule :)

16. anonymous

i see...i left out the 2 for tan^2 mah bad

17. anonymous

a^ u = u(a)^u-1 (u') , sorry lol

18. anonymous

no problem, we all make mistakes ^_^

19. anonymous

i don't really work with sec so my professor said we could leave it in sin or cos format

20. anonymous

but you must know the derivative of tanx = sec^2x while the derivative of secx = secxtanx :)

21. anonymous

which is same as 1/cos^2x

22. anonymous

but I see what you mean :), u left it in cos, it's alright

23. anonymous

yes, true :)

24. anonymous

thank you guys!! ^.^

25. anonymous

you are most welcome sindy ^_^ glad you understood it

26. anonymous

yes i did understand :D ..