Question

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

Answer 1

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`