How do you use logarithmic differentiation to find the derivative of #y=(cosx)^x#?

Question

24203 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-17T13:59:22

#(dy)/(dx)=(cosx)^x[lncosx-xtanx]#

#y=(cosx)^x#

take natural logs of bothe sides

#lny=ln(cosx)^x#

#=>lny=xlncosx#

now differentite #wrt" "x#

the #RHS will need the product rule

#d/(dx)(lny=xlncosx)#

#=>d/(dx)(lny)=lncosxd/(dx)(x)+xd/(dx)(lncosx)#

#1/y(dy)/(dx)=lncosx+x xx (-sinx)/(cosx)#

#(dy)/(dx)=y[lncosx-xtanx]#

substitute for #y#

#(dy)/(dx)=(cosx)^x[lncosx-xtanx]#