How do you find the derivatives of #y=(sintheta)^tantheta# by logarithmic differentiation?

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Answer 1 · 2017-02-10T10:08:27

#(dy)/(dx)=(sintheta)^(tantheta)(sec^2thetalnsintheta+1)#

usinf logarithmic differentiation

#y=(sintheta)^(tantheta)#

#lny=ln(sintheta)^(tantheta)#

by laws of logs

#lny=tanthetaln(sintheta)#

now differentiate #wrt" " x#

#d/dx(lny=tanthetaln(sintheta))#

#RHS" "# will need the product rule

#1/y(dy)/(dx)=sec^2thetalnsintheta+tantheta1/sinthetaxxcostheta#

#1/y(dy)/(dx)=sec^2thetalnsintheta+cancel((sintheta/costhetaxxcostheta/sintheta))^(=1)#

#1/y(dy)/(dx)=sec^2thetalnsintheta+1#

#(dy)/(dx)=y(sec^2thetalnsintheta+1)#

#(dy)/(dx)=(sintheta)^(tantheta)(sec^2thetalnsintheta+1)#