How do you find the derivative of #ln(sqrt(sin(2x)))#?

1 Answer
Guilherme N.
Jan 3, 2016

Chain rule: #(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dw)(dw)/(dx)#

Explanation:

To do so, we'll rename #f(x)=ln(u)#, then #u=sqrt(v)#, #v=sin(w)# and finally #w=2x#

#(dy)/(dx)=1/u(1/(2sqrt(v)))cos(w)(2)=(cancel(2)cos(w))/(cancel(2)usqrt(v))#

Substituting #u#, #v# and #w#, in this order:

#(dy)/(dx)=(cos(w))/(sqrt(v)sqrt(v))=(cos(w))/v=(cos(w))/sin(w)=cos(2x)/sin(2x)=color(green)(cot(2x))#

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