Find an equation of the tangent line to the curve at the given point.

y = sin(sin x), (pi, 0)

1 answer

Nothing simpler...

y = sin(sin x)
y' = cos(sin x)*cos x

y'(pi) = cos(0)*cos(pi) = 1(-1) = -1

so, you want the line through (pi,0) with slope = -1

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