Consider that #(dy)/(dx)=(r'sintheta+rcostheta)/(r'costheta-rsintheta)# where #r'=(dr)/(d theta)#
Since we are given that #r=6cos(theta)#, we can take the derivative of #r# with respect to #theta# to find #r'#:
#r'=-6sin(theta)#
Plugging the equation for #r# and #r'# into the equation for #dy/dx#, we have:
#dy/dx=((-6sintheta)(sintheta)+(6costheta)(costheta))/((-6sintheta)(costheta)-(6costheta)(sintheta))#
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If we are solving for #dy/dx# in general, we can continue to simply this expression:
#dy/dx=(6(cos^2theta-sin^2theta))/(6(-2sinthetacostheta)#
Consider the double-angle formulas: #sin(2theta)=2sinthetacostheta# and #cos(2theta)=cos^2theta-sin^2theta#
Applying these formulas we have:
#dy/dx=-cos(2theta)/sin(2theta)=-cot(2theta)#
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However, since we are given an angle (#theta=(3pi)/2#), if we are actually asked to find #dy/dx|_(theta=(3pi)/2)#, then we can simply use the original plugged in expression:
#dy/dx=((-6sintheta)(sintheta)+(6costheta)(costheta))/((-6sintheta)(costheta)-(6costheta)(sintheta))#
Noting that #cos((3pi)/2)=0#, we can eliminate all terms that have a #costheta#:
#dy/dx|_(theta=(3pi)/2)=((-6sintheta)(sintheta)+0)/(0)#
Notice that we have division by #0# when we do this substitution, this indicates that #dx/(d theta)=0#, so we have a vertical tangent and #dy/dx|_(theta=(3pi)/2)# is undefined.
