## anonymous one year ago How can I find the derivative of tan(5/x) ?

1. IrishBoy123

if you know the derivative of tan u and the derivative of 5/x, you can use the chain rule.

2. IrishBoy123

by letting u = 5/x

3. anonymous

I don't know the derivative of tan u xD Is there any way to find it without having it memorized?

4. SolomonZelman

no need for u

5. IrishBoy123

it's a really good one to know, especially for integration

6. SolomonZelman

$$\large\color{black}{ \displaystyle \frac{d}{dx}\left(~\tan(x/5)~\right) }$$

7. SolomonZelman

how would you differentiate if it was just tan(x) ?

8. anonymous

I don't know lol I get confused when it comes to tan

9. SolomonZelman

what is the derivative of tan(x), you don't know ?

10. anonymous

Nope xD

11. SolomonZelman

you can use a quotient rule, do you know what a quotient rule is?

12. IrishBoy123

@Tracy96 cheers

13. SolomonZelman

Do you know what d/dx means ?

14. anonymous

Ohhh now I remember it was sec^2x hahaha

15. anonymous

Working on math too much today XD lol

16. SolomonZelman

yes, derivative of tan(x) is sec^2(x)

17. SolomonZelman

And when you have tan(x/5) the only difference is that it is sec^2( x/5 ), AND you need to multiply that times the chain rule..... (chain rule for x/5)

18. SolomonZelman

For example, $$\large\color{black}{ \displaystyle \frac{d}{dx}\left(~\tan (4x)~\right)=\sec^2(4x)\times \left(\frac{d}{dx}~4x\right) }$$ $$\large\color{black}{ \displaystyle \frac{d}{dx}\left(~\tan (4x)~\right)=\sec^2(4x)\times ~4 }$$ $$\large\color{black}{ \displaystyle \frac{d}{dx}\left(~\tan (4x)~\right)=4\sec^2(4x) }$$

19. anonymous

Ok so that will be $\sec^2(5/x)(\frac{ -5 }{ x^2 })$

20. SolomonZelman

yes, that is correct, but I would put the last part in front to avoid any confusion

21. SolomonZelman

confusion*

22. anonymous

Right! Thanks so much @SolomonZelman :D

23. SolomonZelman

Sure. Any questions you have regarding this problem or any of the rules?

24. anonymous

Nope! Thanks! That was very helpful :)

25. SolomonZelman

Alright. You are always welcome!

26. anonymous

:)