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

Answer 1

Paisley Johnson

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

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#

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Answer 2

Charles Anctil

#-e^-x#

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|)#

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Answer 3

Samantha Adkison

To differentiate y = e^(-x), you apply the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. Therefore, the derivative of y = e^(-x) is dy/dx = -e^(-x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.