Differentiate $y = x^2e^{-x}$ and find where the tangent line is horizontal.

$y' = xe^{-x}(2 - x)$; horizontal tangents at $(0, 0)$ and $(2, 4e^{-2})$.

A bacteria culture grows as $P(t) = 500 \cdot 2^{\frac{t}{3}}$, where $t$ is in hours. Find the growth rate at $t = 6$, and when the growth rate reaches $1000$ bacteria per hour.

About $462$ bacteria per hour at $t = 6$; the rate reaches $1000$ per hour at $t \approx 9.34$ hours.

Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Derivatives of e^x and a^x

$e^x$ is the one function that is its own derivative: at every point, the slope of $y = e^x$ equals its height. Every other exponential function differentiates to itself times a constant.

How It Works

The two rules.

$\frac{d}{dx}e^x = e^x, \qquad \frac{d}{dx}a^x = a^x \ln a$

The second follows from the first. Since $a = e^{\ln a}$ , you can write $a^x = e^{x\ln a}$ , and the chain rule brings down the constant $\ln a$ . For $a = e$ , $\ln e = 1$ , and the two rules agree.

With the chain rule. When the exponent is anything other than plain $x$ , multiply by the derivative of the exponent:

$\frac{d}{dx}e^{g(x)} = e^{g(x)}g'(x), \qquad \frac{d}{dx}a^{g(x)} = a^{g(x)}\ln a \cdot g'(x)$

So $\frac{d}{dx}e^{3x} = 3e^{3x}$ and $\frac{d}{dx}e^{x^2} = 2xe^{x^2}$ . The exponent stays exactly as it was.

The common mistake is using the power rule on an exponential. The power rule is for a variable base and a constant exponent, like $x^5$ . Here the base is constant and the exponent varies, so $\frac{d}{dx}2^x$ is $2^x \ln 2$ , not $x \cdot 2^{x - 1}$ .

Solving $f'(x) = 0$ . An exponential is never zero, so when a derivative factors as $e^{g(x)}$ times something, only the other factor can be zero. Factor out the exponential and solve what is left. This is how you find the turning points of a function like $x^2e^{-x}$ .

Growth models. For $P(t) = P_0 \cdot b^{\frac{t}{k}}$ , the rate of change is $P'(t) = P_0 \cdot b^{\frac{t}{k}} \cdot \frac{\ln b}{k}$ . The rate is proportional to the amount present, which is what exponential growth means.

Leave $\ln$ values exact unless the question asks for a decimal.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Differentiate $y = 4e^{3x} + x^2$ .
Show solution
1. Chain rule on the exponential: $\frac{d}{dx}4e^{3x} = 4e^{3x} \cdot 3 = 12e^{3x}$ .
2. $y' = 12e^{3x} + 2x$ .
Answer: $y' = 12e^{3x} + 2x$
Easy
Differentiate $y = 3^{2x}$ .
Show solution
1. Base $3$ , exponent $2x$ with derivative $2$ .
2. $y' = 3^{2x} \cdot \ln 3 \cdot 2$ .
Answer: $y' = 2\ln 3 \cdot 3^{2x}$
Medium
Differentiate $y = x^2e^{-x}$ and find where the tangent line is horizontal.
Show solution
1. Product rule: $y' = 2xe^{-x} + x^2 \cdot (-e^{-x})$ .
2. Factor: $y' = xe^{-x}(2 - x)$ .
3. $e^{-x} > 0$ , so $y' = 0$ only when $x = 0$ or $x = 2$ .
4. $y(0) = 0$ and $y(2) = 4e^{-2}$ .
Answer: $y' = xe^{-x}(2 - x)$ ; horizontal tangents at $(0, 0)$ and $(2, 4e^{-2})$ .
Medium
Find the equation of the tangent line to $y = e^{2x - 4}$ at $x = 2$ .
Show solution
1. Point: $y(2) = e^0 = 1$ .
2. $y' = 2e^{2x - 4}$ , so the slope is $2e^0 = 2$ .
3. $y - 1 = 2(x - 2)$ , so $y = 2x - 3$ .
Answer: $y = 2x - 3$
Test level
A bacteria culture grows as $P(t) = 500 \cdot 2^{\frac{t}{3}}$ , where $t$ is in hours. Find the growth rate at $t = 6$ , and when the growth rate reaches $1000$ bacteria per hour.
Show solution
1. $P'(t) = 500 \cdot 2^{\frac{t}{3}} \cdot \ln 2 \cdot \frac{1}{3}$ .
2. $P'(6) = 500 \cdot 4 \cdot \frac{\ln 2}{3} = \frac{2000\ln 2}{3} \approx 462$ bacteria per hour.
3. Set $P'(t) = 1000$ : $2^{\frac{t}{3}} = \frac{6}{\ln 2}$ .
4. Take logarithms: $\frac{t}{3}\ln 2 = \ln\left(\frac{6}{\ln 2}\right)$ , so $t = \frac{3\ln\left(\frac{6}{\ln 2}\right)}{\ln 2} \approx 9.34$ .
Answer: About $462$ bacteria per hour at $t = 6$ ; the rate reaches $1000$ per hour at $t \approx 9.34$ hours.

Learn It on Martian Lab

Martian Lab teaches this in the Derivatives of Logarithmic & Exponential Functions lesson of its MCV4U Derivatives of Trig, Exp and Log unit. Watch it, then take the Derivatives of Trig, Exp and Log practice tests to see if it stuck.

Watch the Derivatives of Logarithmic & Exponential Functions lesson free

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Free Derivatives of Trig, Exp and Log practice test, with solutions

More Derivatives of Trig, Exp and Log topics: Derivatives of Sin, Cos and Tan

More MCV4U topics: Evaluating Limits · Chain Rule · Optimization Problems · Related Rates · Curve Sketching Steps · Dot Product and Projections · Cross Product, Area and Torque · Equations of Planes · Intersection of Three Planes · Continuity · Derivative from First Principles · Power, Product and Quotient Rules · Velocity and Acceleration · Increasing, Decreasing and Critical Points · Concavity and the Second Derivative Test · Adding and Subtracting Vectors · Cartesian Vectors in 2D and 3D · Force and Velocity Vector Problems · Equations of Lines in 3D · Intersection of Lines and Planes · Distance from a Point to a Line or Plane

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