## amyna one year ago y=-3(sec^2x - tan^2(x)) + 4x^3 I do not know how to solve this! Please help with the calc problem!! Thank you!

1. dan815

sec(x) = 1/cos(x)

2. freckles

recall Pythagorean identities $\sin^2(x)+\cos^2(x)=1 \\ \text{ divide both sides by } \cos^2(x) \\ \tan^2(x)+1=\sec^2(x) \\$ anyways I don't know what it is meant by solve

3. freckles

could you say what solve means here @amyna if you are here

4. freckles

you say this is a cal problem do you really mean differentiate are you trying to find the inverse like what are you doing exactly

5. amyna

yup sorry! differentiate it, thanks!

6. freckles

so you need help to differentiate -3+4x^3 just use constant rule and constant multiple rule and power rule?

7. freckles

-3 is a constant the derivative of a constant is 0

8. freckles

$\frac{d}{dx} cx^n \\ c \frac{d}{dx} x^n \text{ by constant multiple rule } \\ c n x^{n-1} \text{ by power rule }$

9. freckles

$\frac{d}{dx}(-3)=? \\ \frac{d}{dx}4x^3=?$

10. amyna

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11. freckles

where did all the sec and tan stuff come from

12. freckles

did you use the Pythagorean identity earlier ? you should y=-3+4x^3

13. freckles

and we are wanting to find y' now

14. freckles

y'=0+12x^2=12x^2

15. freckles

though if you really want to differentiate sec^2(x)-tan^2(x) which is a constant so the derivative should be 0 but pretending you don't know you should wind up with this: $\frac{d}{dx} \sec^2(x)=2 \sec(x) \cdot \sec(x)\tan(x) =2\sec^2(x)\tan(x) \\ \frac{d}{dx} \tan^2(x)=2 \tan(x) \sec^2(x) =2 \sec^2(x)\tan(x) \\ \text{ but as you notice } \\ \frac{d}{dx} \sec^2(x)-\frac{d}{dx} \tan^2(x)=0 \\ \frac{d}{dx}(\sec^2(x)-\tan^2(x))=0$ this is because $\sec^2(x)-\tan^2(x) \text{ is a constant } \\ \text{ this is because } \sec^2(x)-\tan^2(x)=1 \text{ for all } x \text{ on it's domain }$

16. freckles

but as I said all of that work is unnecessary

17. freckles

your y really just say y=-3+4x^3

18. amyna

so what is the final answer?

19. freckles

do you know how to differentiate -3 and 4x^3?

20. amyna

yes its 0 and 12x^2

21. freckles

ok great then why are you asking me what y' is when you know $y=-3+4x^3 \\ y'=0+12x^2 \\ y'=12x^2$

22. amyna

oh okay! i was looking at it wrong! now i get it! thank you so much!

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