To start, we need to determine #f'(x)#, so that we know what the slope of #f(x)# is at any point, #x#.
#f'(x) = d/dx f(x) = d/dx e^x sin^2(x)#
using the product rule:
#f'(x) = (d/dx e^x)sin^2(x) + e^x(d/dx sin^2(x))#
These are standard derivatives:
#d/dx e^x = e^x#
#d/dx sin^2(x) = 2sin(x)cos(x)#
So our derivative becomes:
#f'(x) =e^x sin(x) (sin(x) + 2cos(x))#
Inserting the given #x# value, the slope at #sqrt(pi)# is:
#f'(sqrt(pi)) = e^(sqrt(pi))sin(sqrt(pi)) (sin(sqrt(pi)) + 2cos(sqrt(pi)))#
This is the slope of our line at the point #x= sqrt(pi)#. We can then determine the y intercept by setting:
#y = mx + b#
#m = f'(sqrt(pi))#
#y = f(sqrt(pi))#
This gives us the non-simplified equation for our line:
#f(sqrt(pi)) = (e^(sqrt(pi))sin(sqrt(pi)) (sin(sqrt(pi)) + 2cos(sqrt(pi))))x + b#
#e^(sqrt(pi)) sin^2(sqrt(pi)) = (e^(sqrt(pi))sin(sqrt(pi)) (sin(sqrt(pi)) + 2cos(sqrt(pi))))x + b#
Solving for b, we end up with the annoyingly complicated formula:
#b = e^(sqrt(pi)) sin sqrt(pi)[ sin sqrt(pi) - sqrt(pi)(sin(sqrt(pi)) + 2 cos (sqrt(pi))]#
So our line ends up being:
#y = [e^(sqrt(pi))sin(sqrt(pi)) (sin(sqrt(pi)) + 2cos(sqrt(pi)))]x + e^(sqrt(pi)) sin sqrt(pi)[ sin sqrt(pi) - sqrt(pi)(sin(sqrt(pi)) + 2 cos (sqrt(pi))]#
If we actually calculate what these annoyingly large coefficients equate to, we end up with the approximate line:
#y = 3.34x - 0.27#
