Question

Hi all! How do I take the derivative of this function? (attached)

Answer 1

It's making my head hurt XD

Answer 2

use the log rules,
log A/B = .... ?

Answer 3

and
\(\log \sqrt a = \log a^{1/2}= (1/2)\log a\)

Answer 4

ok I'll try and work through it

Answer 5

is the first step ln(a^2 - z^2)^1/2 - ln(a^2 + z^2)^1/2 ?

Answer 6

yeah, it can be

Answer 7

lol ok

Answer 8

now use this
\(\log \sqrt a = \log a^{1/2}= (1/2)\log a\)

Answer 9

(1/2)ln(a^2 -z^2)-(1/2)ln(a^2 +z^2)

Answer 10

yes, now differentiation will be easier

Answer 11

derivative of lnx is 1/x

Answer 12

should I do that?

Answer 13

yes, and you;ll need chain rule too

Answer 14

f'(g)*g'

Answer 15

yes

Answer 16

ok

Answer 17

(1/2)(1/(a^2 -z^2))(2a-2z)-(1/2)(1/(a^2 +z^2))(2a+2z)

Answer 18

is this right?

Answer 19

'a' is constant, its derivative = 0

Answer 20

aha but z is still some function, right?

Answer 21

yes

Answer 22

so, (1/2)(1/(a^2 -z^2))(-2z)-(1/2)(1/(a^2 +z^2))(2z) ?

Answer 23

yep!

Answer 24

ok so, is the first part -4z/(a^2 -z^2)

Answer 25

yes

Answer 26

overall -4z/(a^2 -z^2) -4z/(a^2 +z^2)

Answer 27

?

Answer 28

seems correct to me

Answer 29

should/can I simplify further?

Answer 30

yes,
-4z can be factored, then take LCM , some terms are getting cancelled...

Answer 31

least common multiple=lcm?

Answer 32

yes

Answer 33

hmm, I think if I factor it out, that should be enough

Answer 34

too much time spent studying math XD

Answer 35

but it can be further simplified...

Answer 36

hmm ok

Answer 37

so I assume I take (a^2 -z^2)/(a^2 -z^2)

Answer 38

?

Answer 39

Well, I have to afk for a bit, then work on this some more. Thanks again as always! :)