What is the equation of the tangent line to the curve #f(x)=6sin(sin(x))# at the point #(pi, 0)#?

1 Answer
Anjali G
Nov 4, 2016

#y=-6(x-pi)#
#y=-6x-6pi#

Explanation:

To find the slope of the tangent line at #(pi,0)#, first find #f'(x)#
#f(x)=6sin(sinx)#
#f'(x)=6(cos(sinx))(cosx)#
#f'(pi)=6(cos(sinpi))(cospi)#
#f'(pi)=6(cos0)(-1)#
#f'(pi)=-6#
This means that the slope of the tangent line at #x = pi# is -6. Now that we have the slope, and we are given a point, to find the equation of the tangent line simply put the information we know into point-slope form:
#y-0=-6(x-pi)#
This simplifies to
#y=-6x-6pi#

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