The slope of the tangent line to a function #f(x)# at #x = x_1# is given by #f'(x_1)#. Recall that the derivative of a function returns its slope at any given point.
We will calculate #f'(x)#. See that we use the product rule.
#f'(x) = e^x secx tanx + e^x secx#
#f'(x) = e^x secx(tanx + 1)#
Evaluating #f'(x)# at #x = pi / 4# gives
#f'(pi/4) = e^(pi/4)sec(pi/4)(tan(pi/4) + 1) = e^(pi/4)(sqrt(2))(1 + 1)#
# = e^(pi/4)(sqrt(2))(2) = e^(pi/4) 2sqrt(2)#
The slope of our tangent line, then, is #m = e^(pi/4) 2sqrt(2)#. We wish to find the line tangent to the point of #f(x)# where #x = pi/4#. This point is given by #(pi/4, f(pi/4)) = (pi/4, e^(pi/4)sqrt(2))#.
Now that we have a slope and a point, we can use slope-intercept form to get our equation.
#y = mx + b#
#y = e^(pi/4) 2sqrt(2) (x) + b#
#e^(pi/4)sqrt(2) = e^(pi/4) 2sqrt(2) (pi/4) + b#
#b = e^(pi/4)sqrt(2) - e^(pi/4)2sqrt(2)(pi/4)#
#b = e^(pi/4)sqrt(2)(1 - (2pi)/4) = e^(pi/4)sqrt(2)(1 - pi/2)#
Having found #b#, we have all the ingredients of our linear equation.
#y = e^(pi/4)2sqrt(2)(x) + e^(pi/4)sqrt(2)(1 - pi/2)#
#y = e^(pi/4)sqrt(2)(2x + 1 - pi/2)#
