What is the equation of the line tangent to #f(x)=x^2 + sin2x # at #x=pi/4#?

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Answer 1 · 2017-10-17T07:29:48

#y=(\pix)/2+(16-\pi^2)/16#

#y=x^2+sin2x=u+v#

#(dy)/(dx)=uv'+vu'#

#u'=2x#

#v'=2cos2x#

#(dy)/(dx)=2xsin2x+2x^2cos2x#

#2(\pi/4)sin(2*\pi/4)+2(\pi/4)^2cos(2*\pi/4)=\pi/2#

#y=(\pi/4)^2+sin(2*\pi/4)=1+\pi^2/16#

#1/1+\pi^2/16=(16+\pi^2)/16#

#y=mx+c#
#(16+\pi^2)/16=\pi/2(\pi/4)+c#

#c=(16+\pi^2)/16-\pi^2/8=(16+\pi^2-2\pi^2)/16=(16-\pi^2)/16#

#y=(\pix)/2+(16-\pi^2)/16#