Question #74cf4

1 Answer
Sayani R.
Oct 21, 2014

y= # (1 + 2/x) ^(2x)#

Taking #log# of both sides,

=> #log# y = #2x# #log (1+ 2/x)#

Differentiating both sides with respect to to x,

=> #1/y dy/dx# = #2x (1/(1+ 2/x)) (-1/x^2) + log (1+2/x) 2 #

=> #dy/dx# = # {-2/( (1+2/x)(x)) + 2 log (1+2/x)} y#

=> #dy/dx# = # {-2/( x+2) + 2 log (1+2/x)} (1 + 2/x)^(2x)#

Related questions
Impact of this question
779 views around the world
You can reuse this answer
Creative Commons License