How do you differentiate #y = log_3 [((x+1)/(x-1))^ln3]#?

Question

2612 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-28T12:07:16

#2/(1-x^2)#

Use #log_ba=ln a/ln b, ln c^n = n ln c and ln(m/s)=ln m - ln n#

Here,

#y=log_3[((x+1)/(x-1))^ln 3]#

#=ln((x+1)/(x-1))^ln 3/ln 3#

#=ln 3 ln[(x+1)/(x-1)]/ln 3#

#=ln(x+1)-ln(x-1)#

y'=1/(x+1)(x+1)'-1/(x-1)(x-1)'#

#=1/(x+1)(1)-1/(x-1)(1)#

#=((x-1)-(x+1))/((x+1)(x-1))#

#=2/(1-x^2)#