Question

How do you differentiate `y=e^-x`?

Answer 1

`d/dx e^-x=-e^-x`

Explanation:

We know that `d/dx a^x=ln(a)*a^x`

`e^-x=(1/e)^x`

Combining the two gives us

`d/dx e^-x=d/dx (1/e)^x=ln(1/e)*(1/e)^x`

`ln(1/e)=ln(e^-1)=-1`

`(1/e)^x=e^-x`

Multiplying the two gives:

`e^-x*-1=-e^-x`

Answer 2

`-e^-x`

Explanation:

Given: `y=e^-x`.

Differentiate using the chain rule, which states that,

`dy/dx=dy/(du)*(du)/dx`

Let `u=-x,:.(du)/dx=-1`.

Then, `y=e^u,dy/(du)=e^u`.

Combine the results together to get:

`dy/dx=e^u*-1`

`=-e^u`

Substitute back `u=-x` to get the final answer:

`color(blue)(bar(ul|=-e^-x|)`