Question #7e44e

1 Answer
Oct 16, 2014

Let #(x_1,y_1)=(pi/2,pi/4)#

By implicitly differentiating the left-hand side with respect to #x#,

#d/{dx}(y sin12x)#

by Product Rule,

#={dy}/{dx}cdot sin12x+y cdot 12cos12x#

#=sin12x{dy}/{dx}+12ycos12x#

by plugging in #(x,y)=(pi/2,pi/4)#,

#=sin(6pi){dy}/{dx}+3picos(6pi)=3pi#

By implicitly differentiating the right-hand side with respect to #x#,

#d/{dx}(x cos2y)#

by Product Rule,

#=1cdot cos2y+x cdot(-sin2y)(2{dy}/{dx})#

#=cos2y-2xsin2y{dy}/{dx}#

by plugging in #(x,y)=(pi/2,pi/4)#,

#=cos(pi/2)-pi sin(pi/2){dy}/{dx}=-pi{dy}/{dx}#

Now, by setting the left-hand side and the right-hand side equal to each other,

#-pi{dy}/{dx}=3pi#

So, the slope #m# of the tangent line is:

#m={dy}/{dx}|_{(pi/2,pi/4)}=-3#

By Point-Slope Form #y-y_1=m(x-x_1)#,

#y-pi/4=-3(x-pi/2)#

I hope that this was helpful.