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)
