Question

Diff >> tan^-1(8/w) =??

Answer 1

u know chain rule ?

Answer 2

U u substitution to solve!

Answer 3

Loi, use chain rule, not sawaken yet ;(

Answer 4

what is d/dx (tan^-1 x) =?

Answer 5

answer is 1/x^2 +1

Answer 6

yes, so what will be d/dw (tan^-1 8/w) ?

Answer 7

just replace x by 8/w

Answer 8

8/w^2 + 64

Answer 9

nopes.
it will be
\(\huge \frac{1}{(8/w)^2+1}.\frac{d}{dw}{(8/w)}\)
this was using chain rule, did u understand this ?

Answer 10

now can u differentiate 8/w ??

Answer 11

d/dw (8/w) = -8/w^2

Answer 12

good :)
so u have
\(\huge \frac{1}{(8/w)^2+1}.\frac{-8}{w^2}\)
can u simplify this?

Answer 13

\[\frac{ d }{ aw } \tan ^{-1}(\frac{ 8 }{ w }) = \frac{ \frac{ -8 }{ e } }{ 1+\frac{ 64 }{ w ^{2} } }\]

Answer 14

yes or no ??

Answer 15

i would rather simplify it like this :
\(\huge \frac{1}{64/w^2+1}.\frac{-8}{w^2}=\frac{-8}{w^2+64}\)
got this ?

Answer 16

Thank you very much.

Answer 17

Welcome very much ^_^

Answer 18

:)

Answer 19

any more questions/doubts?

Answer 20

No questions thank you. :)