Derivative of e^(−x)
A negative exponent changes the sign of the slope. This example separates exponential decay from a negative function value.
Step-by-step solution
Let u=−x. Its derivative is −1.
The derivative of exp(u) with respect to u is exp(u).
f′(x)=exp(−x)·(−1)=−exp(−x).
Domain and tangent behavior
The function is positive for every real x, but its derivative is negative everywhere. It is strictly decreasing. At x=0 the point is (0,1), its slope is −1, and its tangent line is y=1−x.
Second derivative
The second derivative is exp(−x), which is positive. Thus a decreasing function can still be concave up. As x tends to positive infinity, exp(−x) tends to zero without reaching it.
Common mistake
A negative exponent does not make the exponential negative. The minus sign in the derivative comes from differentiating −x.
Practice problems
Differentiate exp(−3x)
Answer: −3exp(−3x): the inner derivative is −3.
Differentiate 5exp(−2x)
Answer: −10exp(−2x): retain the constant 5 and multiply by −2.
Negative exponent versus negative function
The functions e^(−x) and −e^x are different. The first is always positive and represents exponential decay as x increases. The second is always negative and decreases in magnitude as x moves toward negative infinity. For e^(−x), the derivative is negative because the inner exponent −x has derivative −1; the function itself never becomes negative.
Equivalent reciprocal form
Because e^(−x)=1/e^x, the derivative can also be checked with the quotient rule. Differentiating 1/e^x gives [0·e^x−1·e^x]/(e^x)²=−1/e^x=−e^(−x). Reaching the same result through the chain rule and quotient rule is a useful consistency check.
Decay interpretation
The equation f′=−f says the instantaneous rate of change is the negative of the current amount. At x=0 the amount is 1 and the slope is −1. When x increases by ln 2, the value is multiplied by 1/2, so ln 2 is the half-life in this normalized model. This behavior appears in cooling, radioactive decay, and first-order discharge models after units and rate constants are supplied.
More worked variations
e^(−4x): the inner derivative is −4, so f′=−4e^(−4x).
7e^(−x): the constant multiple remains, producing −7e^(−x).
e^(1−x²): the inner derivative is −2x, giving −2x e^(1−x²). The constant 1 does not contribute.
x e^(−x): product and chain rules give e^(−x)−x e^(−x)=(1−x)e^(−x).
Limits and concavity
As x approaches positive infinity, e^(−x) approaches zero from above but never reaches zero. As x approaches negative infinity, it grows without bound. Its second derivative is e^(−x), which is positive everywhere, so the decreasing graph is concave up across its full real domain.
Questions students ask
Why does the second derivative become positive?
Differentiating −e^(−x) produces two negative factors: the existing minus sign and the inner derivative −1.
Does the horizontal axis belong to the graph?
No. y=0 is a horizontal asymptote as x increases, but e^(−x) remains strictly positive for every finite x.
Related exponential derivatives
Review the chain rule and product rule, or read our verification method and limitations.