How do you differentiate #f(x)=x-sqrt(2^x-x^2)# using the chain rule?

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Answer 1 · 2015-12-02T23:27:06

#f'(x)=1-(2^xln2-2x)/(2sqrt(2^x-x^2))#

#f'(x)=1-d/dx[(2^x-x^2)^(1/2)]#

Find just this derivative through the Chain Rule:

#d/dx[(2^x-x^2)^(1/2)]=1/2(2^x-x^2)^(-1/2)d/dx[2^x-x^2]#

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

Find the remaining derivative.

#d/dx[2^x]=d/dx[e^(ln2^x)]=d/dx[e^(xln2)]=e^(xln2)d/dx[xln2]=2^xln2#

Plug back in:

#d/dx[(2^x-x^2)^(1/2)]=(2^xln2-2x)/(2sqrt(2^x-x^2))#

Plug in again.

#f'(x)=1-(2^xln2-2x)/(2sqrt(2^x-x^2))#

(Optional simplification:)

#f'(x)=(2sqrt(2^x-x^2)-2^xln2+2x)/(2sqrt(2^x-x^2))#