What is the equation of the tangent line of #f(x)=x^2 + sinx # at #x=pi#?

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Answer 1 · 2015-12-03T02:40:36

#y=(2pi-1)x-pi^2+pi#

Find #f'(pi)# at the point #(pi,pi^2)#.

#f'(x)=2x+cosx#

#f'(pi)=2pi+cospi#

#f'(pi)=2pi-1#

Thus, #2pi-1# is the slope of the tangent line at the point #(pi,pi^2)#.

Write the equation of the line in point-slope form:

#y-pi^2=(2pi-1)(x-pi)#

In slope-intercept form: #y=(2pi-1)x-pi^2+pi#