We nave f(x)=e^(-x^2+x+pi)-sinx/e^xf(x)=e−x2+x+π−sinxex
The equation of the tangent will be in the form of y=mx+cy=mx+c
First, we will find the gradient mm:
f(x)=e^(-x^2+x+pi)-sinx/e^xf(x)=e−x2+x+π−sinxex
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