Answer:
Equation of tangent to the line will be [tex]2e^2x-4e^2-2x+5[/tex]
Step-by-step explanation:
We have given the equation of the line [tex]y=e^2-x[/tex]
Now slope of tangent will be equal to [tex]\frac{dy}{dx}=e^2x-\frac{x^2}{2}[/tex]
Slope at point (2,1)
So [tex]\frac{dy}{dx}=e^2\times 2-\frac{2^2}{2}=2e^2-2[/tex]
Now equation of line is given as [tex]y-y_1=m(x-x_!)[/tex]
So [tex]y-1=(2e^2-2)(x-2)[/tex]
[tex]y=(2e^2-2)(x-2)+1=2e^2x-4e^2-2x+4+1=2e^2x-4e^2-2x+5[/tex]
