How do you differentiate # f(x)=sqrt(ln(x^2+3)# using the chain rule.?

1 Answer
1s2s2p
Apr 9, 2018

#f'(x)=(x(ln(x^2+3))^(-1/2))/(x^2+3)=x/((x^2+3)(ln(x^2+3))^(1/2))=x/((x^2+3)sqrt(ln(x^2+3)))#

Explanation:

We are given:
#y=(ln(x^2+3))^(1/2)#
#y'=1/2*(ln(x^2+3))^(1/2-1)*d/dx[ln(x^2+3)]#
#y'=(ln(x^2+3))^(-1/2)/2*d/dx[ln(x^2+3)]#

#d/dx[ln(x^2+3)]=(d/dx[x^2+3])/(x^2+3)#

#d/dx[x^2+3]=2x#

#y'=(ln(x^2+3))^(-1/2)/2*(2x)/(x^2+3)=(x(ln(x^2+3))^(-1/2))/(x^2+3)=x/((x^2+3)(ln(x^2+3))^(1/2))=x/((x^2+3)sqrt(ln(x^2+3)))#

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