How do you differentiate #f(x)=sqrt(cote^(4x)# using the chain rule.?

Question

1255 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-05T12:55:54

#f'(x)=(-4e^(4x)csc^2(e^(4x))(cot(e^(4x)))^(-1/2))/2#
#color(white)(f'(x))=-(2e^(4x)csc^2(e^(4x)))/sqrt(cot(e^(4x))#

#f(x)=sqrt(cot(e^(4x)))#
#color(white)(f(x))=sqrt(g(x))#
#f'(x)=1/2*(g(x))^(-1/2)*g'(x)#
#color(white)(f'(x))=(g'(x)(g(x))^(-1/2))/2#

#g(x)=cot(e^(4x))#
#color(white)(g(x))=cot(h(x))#
#g'(x)=-h'(x)csc^2(h(x))#

#h(x)=e^(4x)#
#color(white)(h(x))=e^(j(x))#
#h'(x)=j'(x)e^(j(x))#

#j(x)=4x#
#j'(x)=4#

#h'(x)=4e^(4x)#

#g'(x)=-4e^(4x)csc^2(e^(4x))#

#f'(x)=(-4e^(4x)csc^2(e^(4x))(cot(e^(4x)))^(-1/2))/2#
#color(white)(f'(x))=-(2e^(4x)csc^2(e^(4x)))/sqrt(cot(e^(4x))#