Question #5d4ec

1 Answer
Oct 7, 2015

#y=x#

Explanation:

#y=sinx+sin^2x#

#(dy)/(dx)=cosx+2sinxcosx#

At #x=0#,

#(dy)/(dx)=cos(0)+2sin(0)cos(0)#

#(dy)/(dx)=1#

Plugging #y_1=0#, #x_1=0#, and #(dy)/(dx)=1# into the general equation for a line, #y-y_1=m(x-x_1)#, we get

#y-0=1(x-0)#

#y=x#