How do you use implicit differentiation to find an equation of the tangent line to the curve at the given point #ysin16x=xcos2y# and #(pi/2, pi/4)#?

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Answer 1 · 2017-05-04T08:23:55

#y+4x=(9pi)/4#

The slope of the tangent to any curve is given by #(dy)/(dx)#

As #ysin16x=xcos2y# and differentiating

#(dy)/(dx)sin16x+16ycos16x=1xxcos2y-2xsin2y(dy)/(dx)#

or #(dy)/(dx)sin16x+2xsin2y(dy)/(dx)=cos2y-16ycos16x#

or #(dy)/(dx)=(cos2y-16ycos16x)/(sin16x+2xsin2y)#

Hence at #(pi/2,pi/4)# slope is

#(cos2(pi/4)-16(pi/4)cos(4pi))/(sin(8pi)+2pi/2sin(pi/2))#

= #(0-4pi)/(0+pi)=-4#

As tangent is at #(pi/2,pi/4)#, its equation is

#(y-pi/4)=-4(x-pi/2)# or #y+4x=(9pi)/4#