What is the equation of the line tangent to #f(x)=x ^2cos^2(x) # at #x=pi/4#?

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Answer 1 · 2018-06-02T08:38:59

See below

The equation of tangent line is given by

(1) #(y-y_0)=f´(x_0)(x-x_0)# where #x_0=pi/4# and #y_0=f(x_0)# and #f´(x_0)# is the value of derivative of #f# in #x_0#

Proceed:

#x_0=pi/4#
#f(pi/4)=pi^2/16cos^2(pi/4)=pi^2/32=y_0#

#f´(x)=2xcos^2x-2x^2cosxsinx=2xcos^2x-x^2sin2x#

#f´(pi/4)=cancel2·pi/4·1/cancel2-pi^2/16·1=pi/4-pi^2/16#

Apply results in (1)

#(y-pi^2/16)=(pi/4-pi^2/16)(x-pi/4)# is the tangent line equation requested