How to differentiate #y = e^sqrt(1-x)#. And how to eliminate e from the equation?

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Answer 1 · 2018-05-19T10:50:46

#(dy)/(dx)=(-y)/(2lny)#

We know that,

"If #a in RR^+ -{1} , x inRR and y in RR^+, then#

#y=a^x<=>log_ay=x# "

So,

#y=e^X<=>lny=X...to(1)#

Using #(1)# we get

#y = e^sqrt(1-x)#

#=>lny=sqrt(1-x)...to(A)#

Diff.w.r.t. #x#

#1/y(dy)/(dx)=1/(2sqrt(1-x))xx(-1)#

#=>(dy)/(dx)=(-y)/(2sqrt(1-x))#

From #(A)#,we have #sqrt(1-x)=lny#

#=>(dy)/(dx)=(-y)/(2lny)#