How do you differentiate # y = sin (ln(x)^2+2)# using the chain rule?

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Answer 1 · 2018-07-30T09:45:15

#(dy)/(dx)=2/xlnx cos(ln(x)^2+2)#

Here ,

#y=sin(ln(x)^2+2)#

Let , # y=sinu# , where, #color(red)(u=ln(x)^2+2#

#=>(dy)/(du)=cosu and (du)/(dx)=2lnx*1/x=2/xlnx#

#color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)#

#:.(dy)/(dx)=cosu xx 2/xlnx=2/xlnxcosu#

Subst. #color(red)(u=ln(x)^2+2#

#:.(dy)/(dx)=2/xlnx cos(ln(x)^2+2)#