## anonymous one year ago Can someone show me how to find d/dx (arctan (5x))

1. UsukiDoll

$\frac{d}{dx} \tan^{-1}(5x)$ this?

2. UsukiDoll

ok. the formula for the derivative of the inverse trigonometric function is $\tan^{-1}(x) = \frac{1}{1+x^2}$

3. UsukiDoll

so we need to find the derivative of 5x first what is the derivative of 5x? @cstarxq

4. UsukiDoll

I'm not using the right variable here...this is going to be confusing a bit $\frac{d}{dy}$$\tan^{-1}(y) = \frac{1}{1+y^2}$ let y = 5x so we need the derivative of 5x which is?

5. anonymous

5

6. UsukiDoll

mhm that's correct so that goes in the front and then afterwards you just substitute y=5x in the fraction

7. UsukiDoll

$\frac{5}{1+y^2}$ so if y =5x then we have (5x)^2 which is

8. UsukiDoll

(5x)^2 = (5x)(5x) = ?

9. anonymous

25x^2

10. UsukiDoll

correct and that's it.. $\frac{5}{1+25x^2}$

11. UsukiDoll

you just have to take the derivative of the inside for $\frac{d}{dx} \tan^{-1}(5x)$ in that case it was a 5x so the derivative of that is 5

12. anonymous

oh i see, i forgot i had to take the derivative of 5x and substituting that into y^2..

13. UsukiDoll

then the derivative goes on the numerator.. and then the denominator is just pure substitution

14. UsukiDoll

can I get a medal ? ^^

15. UsukiDoll

thank you :D

16. anonymous

17. UsukiDoll

you're welcome :) If I remembered correctly, you need to memorize them >.< Calculus II had so much formulas.

18. anonymous

x____x oh, yeah it does

19. UsukiDoll

Calculus III should be easy, but Calculus IV... UGH! So many conversations, integrations, and vectors come back

20. anonymous

I lost my connection, but I'm taking calc III and linear algebra next semester. Hopefully it won't be too bad.

21. UsukiDoll

22. idku

whether you are still viewing or not, there is a way of doing it. Don't be a slave of a non-making sense (or even making sense) formula. Every formula you use should be something you understand and that is what I will try to improve right now.

23. idku

you are dealing with an inverse function. Let's say inv. tan. $$y={\rm Tan}^{-1}(x)$$ re-write it first. $$x={\rm Tan}^{}(y)$$ (i hope you know why I did this, or elsego back and rvw that rule) Now differentiate with respect to x. (Taking a derivative of a function y=g(x), that is dy/dx) $$\frac{dy}{dx}~x=\frac{dy}{dx}~{\rm Tan}^{}(y)$$ $$1={\rm Sec}^{2}(y)\times y'$$ $$1/{\rm Sec}^{2}(y)= y'$$ $$1/\left({\rm Tan}^{2}(y)+1\right) = y'$$ you know what tan(y) is from the beginning, and thus you know what tan^2(y) is as well. $$1/\left(x^2+1\right) = y'$$ and there is your derivative

24. idku

same you can do with any inverse function.....

25. anonymous

Yes, I'm still reviewing this. Thanks @idku, I just used your method to evaluate the derivative for the original problem I was trying to figure out which was actually using integration by parts to evaluate integral of arctan(5x)...