How do you differentiate #y=sqrt(2-e^x)#?

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Answer 1 · 2017-04-04T15:32:07

Use the chain rule:
#(d(f(g(x))))/dx = (df(g))/(dg)(d(g(x)))/dx#

Let #g(x) = 2 - e^x# then it follows that:

#(d(g(x)))/dx= -e^x#

#f(g) = sqrt(g)#

and

#(df(g))/(dg) = 1/(2sqrtg)#

Substituting this into the chain rule:

#(d(sqrt(2-e^x)))/dx =1/(2sqrtg)-e^x#

#(d(sqrt(2-e^x)))/dx =-e^x/(2sqrtg)#

Reverse the substitution for g:

#(d(sqrt(2-e^x)))/dx =-e^x/(2sqrt(2-e^x))#