## ParthKohli 3 years ago An easy way to memorize the differentiations of trig functions?

1. klimenkov

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2. ParthKohli

Can this kinda chain help? sin(x) | cos(x) | -sin(x) | -cos(x) | SIN(X)

3. etemplin

there is no easy way. it comes with enough practice and problems. Thats how i got them. sin=cos, cos=-sin, tan = sec^2, sec = sec*tan...thats what i know...the other 2 dont get used that often. the chain/circular reference does kinda work

4. ParthKohli

Yeah, but where's the tan(x)?

5. ParthKohli

And the hyperbolic ones?

6. ParthKohli

cos = -sin ^

7. etemplin

ive never had to use the hyperbolic ones, and the inverse ones (i.e. sin^-1) are very difficult to memorize because they all look the same

8. ParthKohli

Hmm... okay.

9. ParthKohli

But still, is there a way to memorize all of this stuff?

10. etemplin

do lots of problems. Eventually you'll memorize them because you see them so often

11. ParthKohli

How about the sec and cosec?

12. etemplin

sec = sec*tan, csc = -cscx*cotx. theyre basically opposites, so if you know sec, you can get csc

13. klimenkov

You can refer to such thing, bind sin and cos, tan and sec, cot and cosec. They are together in different formulas.

14. NickR

yup practice

15. zordoloom

I would really like to know this too. Memorizing diff trig variables.

16. lgbasallote

just remember: derivative of a cofunction is negative i.e. d/dx of cos x is -sinx d/dx of csc x is -csc x cot x d/dx of cot x is -csc^2 x

17. NickR

would help alot actually

18. lgbasallote

and most importantly: hate math

19. calculusfunctions

I've never had to memorize them. After doing them a few times, you just know them. I'm only speaking for myself, but I just know all the differentiation rules, trig identities by heart. Actually pretty much any formula, I've never had to really memorize, but rather just understand the justification behind them. I can't explain it but I pretty much know any mathematical formula, rules, etc, by heart. I guess that's why I love MATH!

20. ParthKohli

Hmm. I'd try that way. Thanks @calculusfunctions