Let f(x) = e^x cos(x)
then the gradient of f(x) is the derivative of y wrt x or
f'(x) = e^x(cos(x) - sin(x))
When x = pi, the gradient becomes
f'(pi) = e^pi(cos(pi) - sin(pi)) = -e^pi
Since the product of the gradients of two perpendicular lines is equal to -1, the gradient of the normal to the curve is equal to -1/f'(x)
Therefore the gradient of the normal to the curve at the point x = pi is
-1 / -e^pi = e^-pi