Question

Can someone Help me with this pls!
H(x)=ln.sqrt(a^2 -z^2/a^2 +z^2)

Answer 1

what the question is?

Answer 2

Differentiate the function!

Answer 3

partial derivative?

Answer 4

I guess Inverse functions

Answer 5

which class are you in? cal1,2,3?

Answer 6

cal-2

Answer 7

ok, to me, I break the ln and sqr into 2 parts. I have H(x) = ln (sqr(a^2 -z^2) ) - ln (sqr(a^2 + z^2) ). then take derivative each of them. don't forget each is 3 layers function

Answer 8

at cal2, you just have 1 variable, I assume that your a is constant, so, just take derivative respect to z

Answer 9

HOW CAN YOU BREAK THE sqrt into 2?

Answer 10

I mean, is it possible?

Answer 11

\[\sqrt{\frac{ a^2-z^2 }{ a^2+z^2 }}= \frac{ \sqrt{a^2-z^2} }{ \sqrt{a^2+z^2} }\]

Answer 12

is it ok?

Answer 13

Thanks man :) btw what do U mean by 3 layers function?

Answer 14

you have ln-sqrroot- sqr . that is 3 layers

Answer 15

when take derivative, go from outside to inside, i mean from left to right.

Answer 16

unfan me please, I just stop by for fun,not staying here long, try to become a gentlement

Answer 17

OK! but I still didn't get the case"3 layers functions"! becoz we have ln & sqrt there so it must be 2 layers!!!

Answer 18

hey, how about z^2?

Answer 19

when you "touch" it, you must take (z^2)' to get ... something, I don't know

Answer 20

it's inside the root , so the whole function under the root must be one layer itself!
am I wrong?

Answer 21

unfortunately, you are not right. if z is just z, you are right, but z is z^2 , you must consider it a function of z

Answer 22

So ln, a^2, z^2 each considered as a function here?!!
btw thx in advance ;)

Answer 23

I mean as a layer!

Answer 24

1 personal question: are you taking cal-3? which university?

Answer 25

I do the first part for you [ln(sqr)(a^2 -z^2 ]= [1/sqr (a^2-z^2) ]*[ (sqr(a^2 -z^2)]' .
the second part of this part ask you take derivative of sqr, right? I break it again to count just that part. you must time to the first part then.
[sqr(a^2-z^2)]' = 1/2sqr(a^2-z^2) * (a^2-z^2)' . and (a^2-z^2)' = 2z. is it right?

Answer 26

yeap. not university, just college. community college of Philadelphia

Answer 27

well, thank you so much for your help :) by the way you look one of those timid girls who are running away from guys, aren't you? (bcoz U asked me to unfollow you!)

Answer 28

nope, I am a gentlemen. I do it because there are so many people become my fan and waiting for help when they need help while I don't have time for them. I don't want you become one of them, ask for help and then wait hopelessly.

Answer 29

I have to go now, if I see you when you need help AND I have time, definitely I am there. bye bye

Answer 30

yeah, I see! have a good time (it is night here but it must be afternoon there...! i usually come for questions at nights) ;) bye now