How do you find the equation of the tangent line #y=e^-x# at #(0,1)#?

Question

18226 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-26T23:46:20

#y=-x+1#

One of the amazing thing with #e# is that:

#d/dx(e^x)=e^x*d/dx(x)=>e^x#

Using this, we have:

#d/dx(y)=d/dx(e^-x)#

#dy/dx=e^(-x)*d/dx(-x)#

We use the power rule here:

#x^n=nx^(n-1)# when #n# is a constant.

#dy/dx=e^(-x)*-1#

#dy/dx=-e^-x#

When #x=0#,

#m=-e^-0#

#m=-1#

Using #y-y_1=m(x-x_1)#,

#y-1=-1(x-0)#

#y=-x+0+1#

#y=-x+1#