What is the equation of the tangent line of f(x)=e^(-x^2+x+pi)-sinx/e^xf(x)=ex2+x+πsinxex at x=pi/4x=π4?

1 Answer
Jan 18, 2018

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)y=(1π2)e5π4π216x(π2+2π8)e5π4π2168eπ42
y~~-15.633526x+39.3451202y15.633526x+39.3451202
y~~-15.6x+39.3y15.6x+39.3

Explanation:

We nave f(x)=e^(-x^2+x+pi)-sinx/e^xf(x)=ex2+x+πsinxex

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)=ex2+x+πsinxex

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