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

1 Answer
May 29, 2016

Equation of tangent is #6.737x-y-2.671=0#

Explanation:

At #x=pi/4#, #f(x)=e^((pi/4)^2)/cos(pi/4)=1.853082/0.707107=2.62#

Hence the tangent passes through #(pi/4,2.62)#

Slope of the tangent is given by #f'(x)#

As #f'(x)=(cosx*e^(x^2)*2x-e^(x^2)(-sinx))/cos^2x#

and at #x=pi/4#, it is

#f'(pi/4)=(cos(pi/4)*e^((pi/4)^2)*2(pi/4)+sin(pi/4)e^((pi/4)^2))/cos^2x#

= #(1/sqrt2*e^(pi^2/16)*(pi/2)+1/sqrt2e^(pi^2/16))/(1/2)#

= #sqrt2*e^(pi^2/16)*(pi/2)+sqrt2e^(pi^2/16)#

= #1.4142*1.853*1.5708+1.4142*1.853=6.737#

As such equation of tangent is #y-2.62=6.737(x-0.7854)#

#6.737x-y+2.62-5.291=0# or #6.737x-y-2.671=0#