We nave #f(x)=e^(-x^2+x+pi)-sinx/e^x#
The equation of the tangent will be in the form of #y=mx+c#
First, we will find the gradient #m#:
#f(x)=e^(-x^2+x+pi)-sinx/e^x#
#f'(x)=d/dx[e^(-x^2+x+pi)-sinx/e^x]#
#color(white)(f'(x))=d/dx[e^(-x^2+x+pi)]-d/dx[sinx/e^x]#
#color(white)(f'(x))=d/dx[-x^2+x+pi]e^(-x^2+x+pi)-(e^xd/dx[sinx]-sinxd/dx[e^x])/(e^x)^2#
#color(white)(f'(x))=(-2x+1)e^(-x^2+x+pi)-(e^xcosx-sinxe^x)/(e^x)^2#
#color(white)(f'(x))=(-2x+1)e^(-x^2+x+pi)-(cosx-sinx)/e^x#
#f'(pi/4)=(-2(pi/4)+1)e^(-(pi/4)^2+(pi/4)+pi)-(cos(pi/4)-sin(pi/4))/e^(pi/4)#
#color(white)(f'(pi/4))=(1-pi/2)e^((5pi)/4-(pi^2)/16)#
Now for #x# and #y#
We know #x=pi/4#
Though #y=f(pi/4)=e^(-(pi/4)^2+(pi/4)+pi)-sin(pi/4)/e^(pi/4)=e^((5*pi)/4-pi^2/16)-e^(-pi/4)/sqrt(2)#
Finally, #c#:
#y=mx+c#
#c=y-mx#
#color(white)(c)=(e^((5*pi)/4-pi^2/16)-e^(-pi/4)/sqrt(2))-pi/4((1-pi/2)e^((5pi)/4-(pi^2)/16))#
#color(white)(c)-((-pi^2+2*pi-8)*e^((5*pi)/4-pi^2/16))/8-e^(-pi/4)/sqrt(2)#
Overall, #y=(1-pi/2)e^((5pi)/4-(pi^2)/16)x-((-pi^2+2*pi-8)*e^((5*pi)/4-pi^2/16))/8-e^(-pi/4)/sqrt(2)~~-15.633526x+39.3451202~~-15.6x+39.3#
