How do you find the equation of the tangent line to the curve #y=sin(sinx)# at (pi,0)?

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Answer 1 · 2018-04-12T22:02:26

#y = pi- x#

If we take the derivative, we get by the chain rule that

#y' = cos(sinx)cosx#

At #x = pi# this takes the value of

#y'(pi) = cos(sinpi)cos(pi) = cos(0)(-1) = -1#

Therefore the equation of the tangent is

#y -y_1 = m(x- x_1)#

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

#y = -x + pi#

Hopefully this helps!