09/03/2026

#Math#Differentiation

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CS50P Lecture 0: Functions, Variables

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Exercise 5 PDF

The Exponential Function $e^{x}$

Recitation 4 previewed the pattern that every tangent slope on $y = b^{x}$ is a fixed multiple of the slope at $x = 0$ . The 15% investment, the doubling bacteria culture, and the decaying $^{235}\mathrm{U}$ pile all share the proportionality $f' = k \, f$ , and the only quantity left to pin down is the slope of $b^{x}$ at $x = 0$ . Once that slope is named, the derivative is forced everywhere.

Three differentiation rules from Lesson 4AM stand ready: the product rule, the quotient rule, and the chain rule. None of them apply to $b^{x}$ as written, because $b^{x}$ is not a polynomial, not a quotient of polynomials, and not a composition involving such pieces. The limit definition of the derivative from Lesson 2PM is the only tool that reaches it.

The Slope of $2^{x}$ at $x = 0$

Denote the slope of $y = 2^{x}$ at $x = 0$ by $m$ . By the limit definition from Lesson 2PM, applied at $x = 0$ ,

m = \lim_{h \to 0} \frac{2^{0 + h} - 2^{0}}{h} = \lim_{h \to 0} \frac{2^{h} - 1}{h}. \tag{1}

Each value of $h$ produces a secant line through $(0, 1)$ and $(h, 2^{h})$ , with slope $(2^{h} - 1)/h$ ; as $h \to 0$ , the secant tilts toward the tangent.

On the curve y = 2^x, the secant line through (0, 1) and (h, 2^h) for h = 0.7 is dashed, sloping more steeply than the true tangent at x = 0 drawn solid. As h shrinks toward zero, the secant tilts down toward the tangent.

The right-hand side of $(1)$ has the indeterminate form $0/0$ at $h = 0$ , and the algebraic moves available, namely the laws of exponents from Recitation 4, give $2^{h} - 1$ no useful factorisation. The limit must be estimated numerically. Tabulating $(2^{h} - 1)/h$ for $h = 0.1, 0.01, \dots, 10^{-7}$ :

$h$	$(2^{h} - 1)/h$
$0.1$	$0.71773$
$0.01$	$0.69556$
$0.001$	$0.69339$
$0.0001$	$0.69317$
$10^{-5}$	$0.69315$
$10^{-6}$	$0.69315$
$10^{-7}$	$0.69315$

The values stabilise at $0.69315\dots$ to five decimal places. The full proof that the limiting process converges belongs to the later analysis course; here we record the limiting value supplied by the table:

m = \frac{d}{dx} (2^{x}) \!\bigg|_{x = 0} \approx 0.693. \tag{2}

The same construction performed for $b = 3$ would tabulate $(3^{h} - 1)/h$ and stabilise near $1.0986$ ; for $b = 5$ , near $1.6094$ . These values match the qualitative ranking visible in Recitation 4’s five-curve figure: larger bases give steeper exponential curves near $x = 0$ .

The Derivative of $2^{x}$ at Every $x$

The slope at $x = 0$ propagates to every $x$ by Recitation 4’s structural observation. To turn that into a proof, run the limit definition at an arbitrary $x$ and use law (i) of Recitation 4 to peel off the $x$ .

By the limit definition,

\frac{d}{dx} (2^{x}) = \lim_{h \to 0} \frac{2^{x + h} - 2^{x}}{h}. \tag{3}

Law (i) of exponents from Recitation 4 gives $2^{x + h} = 2^{x} \cdot 2^{h}$ , so

2^{x + h} - 2^{x} = 2^{x} (2^{h} - 1).

Substituting into $(3)$ and pulling the constant $2^{x}$ outside the limit (the variable of the limit is $h$ , not $x$ ):

\frac{d}{dx} (2^{x}) = \lim_{h \to 0} 2^{x} \cdot \frac{2^{h} - 1}{h} = 2^{x} \cdot \lim_{h \to 0} \frac{2^{h} - 1}{h} = m \cdot 2^{x}. \tag{4}

The unknown $m$ is the same constant from $(2)$ , the slope at zero, and it factors out of every other derivative computation involving $2^{x}$ . Compactly,

\frac{d}{dx} (2^{x}) = m \cdot 2^{x}, \qquad m \approx 0.693. \tag{5}

Example 130 (Slopes of

2^{x}

at sample points)

Compute:

$\dfrac{d}{dx}(2^{x}) \!\bigg|_{x = 3}$ .
$\dfrac{d}{dx}(2^{x}) \!\bigg|_{x = -1}$ .

By $(5)$ ,

\frac{d}{dx}(2^{x}) \!\bigg|_{x = 3} = m \cdot 2^{3} = 8 m \approx 8 (0.693) = 5.544,

\frac{d}{dx}(2^{x}) \!\bigg|_{x = -1} = m \cdot 2^{-1} = 0.5 \, m \approx 0.5 (0.693) = 0.347.

The derivative eight units to the right is sixteen times the derivative one unit to the left, because $2^{3} / 2^{-1} = 2^{4} = 16$ . The function and its derivative grow in lockstep.

The argument that produced $(5)$ used only law (i) of Recitation 4, the exponential add-then-multiply identity. Replacing $2$ by any base $b > 0$ leaves every step intact:

\frac{d}{dx} (b^{x}) = m_{b} \cdot b^{x}, \qquad m_{b} = \frac{d}{dx} (b^{x}) \!\bigg|_{x = 0} = \lim_{h \to 0} \frac{b^{h} - 1}{h}. \tag{6}

The slope at zero is base-dependent; the form of the derivative, function multiplied by a constant, is not.

Problem 141

Use $(6)$ together with the numerical values $m_{2} \approx 0.693$ , $m_{3} \approx 1.099$ , and $m_{5} \approx 1.609$ to compute each derivative.

$\dfrac{d}{dx}(3^{x}) \!\bigg|_{x = 2}$ .
$\dfrac{d}{dx}(5^{x}) \!\bigg|_{x = 0}$ and $\dfrac{d}{dx}(5^{x}) \!\bigg|_{x = 1}$ .
The ratio $\dfrac{(d/dx)(2^{x})|_{x = 4}}{(d/dx)(2^{x})|_{x = 1}}$ , simplified by cancellation of $m_{2}$ before any numerical work.

Problem 142

Tabulate $(3^{h} - 1)/h$ at $h = 0.1, 0.01, 0.001$ to confirm $m_{3} \approx 1.099$ to three decimal places. Compare with the value of $m_{3}$ suggested by Recitation 4’s five-curve figure.

The Number $e$

Equation $(6)$ is awkward because every base carries its own undetermined constant $m_{b}$ . The numerical experiments above show $m_{2} \approx 0.693 < 1$ and $m_{3} \approx 1.099 > 1$ . More detailed tables show that exactly one base between $2$ and $3$ has slope $1$ at $x = 0$ . A full proof of this existence and uniqueness is part of real analysis; in MA0A we take it as the defining fact of the number below.

Definition 36 (The number

e

)

The number $e$ is the unique base $b > 0$ for which

\lim_{h \to 0} \frac{b^{h} - 1}{h} = 1,

equivalently, the unique $b$ for which the graph of $y = b^{x}$ has slope exactly $1$ at $x = 0$ .

To ten significant digits, $e = 2.718281828\dots$ . For most purposes, $e \approx 2.7$ suffices.

Substituting $b = e$ into $(6)$ collapses the formula:

\frac{d}{dx} (e^{x}) = m_{e} \cdot e^{x} = 1 \cdot e^{x} = e^{x}. \tag{7}

Theorem 17 (Derivative of

e^{x}

)

For every real $x$ ,

\frac{d}{dx} (e^{x}) = e^{x}.

Equivalently, the function $f(x) = e^{x}$ satisfies $f' = f$ .

Geometrically, at every point on the curve $y = e^{x}$ the slope of the tangent line equals the height of the curve.

The curve y = e^x with four tangent lines drawn at x = -1, 0, 1, 2. The slopes of the tangents are 1/e, 1, e, and e^2 respectively, exactly the heights of the curve at those x-values.

At $x = -1$ , the height $1/e \approx 0.37$ equals the slope. At $x = 0$ , the height $1$ equals the slope. At $x = 1$ , the height $e$ equals the slope. At $x = 2$ , the height $e^{2}$ equals the slope.

Example 131 (The tangent line at

x = 1

)

Find the tangent line to $y = e^{x}$ at $x = 1$ .

The point of tangency is $(1, e)$ . The slope is $e^{1} = e$ . Point-slope form gives

y - e = e (x - 1),

so $y = e x$ .

The tangent line passes through the origin with slope $e$ . In rate-of-change language, near $x = 1$ a small input change $\Delta x$ changes the output by approximately $e\,\Delta x$ , and the tangent line is the linear approximation at that point.

Problem 143

Find the tangent line to $y = e^{x}$ at $x = 0$ and at $x = -1$ . In each case, state where the tangent line meets the $x$ -axis.

Problem 144

Show that the tangent line to $y = e^{x}$ at the point $(a, e^{a})$ has $x$ -intercept $a - 1$ . Conclude that, as $a$ varies, the foot of the tangent line slides along the $x$ -axis at the same rate as $a$ itself moves.

$e^{x}$ in the Toolkit

The derivative formula for $e^{x}$ , combined with the product, quotient, and chain rules from Lesson 4AM, differentiates every expression built from $e^{x}$ and the polynomial/rational pieces of earlier lessons.

Example 132 (Algebra of exponents with base

e

)

Write each expression in the form $A \, e^{k x}$ for constants $A$ and $k$ .

$\dfrac{(3 e^{2 x})^{2} \cdot e^{x}}{e^{-2 x}}$ .
$\sqrt{\dfrac{e^{x}}{e^{7 x}}} \cdot e^{2 x + 5}$ .

The laws of exponents from Recitation 4 transfer unchanged to base $e$ .

Square the inside, then collect:

\frac{(3 e^{2 x})^{2} \cdot e^{x}}{e^{-2 x}} = \frac{9 \, e^{4 x} \cdot e^{x}}{e^{-2 x}} = 9 \, e^{4 x} \cdot e^{x} \cdot e^{2 x} = 9 \, e^{4 x + x + 2 x} = 9 \, e^{7 x}.

So $A = 9$ , $k = 7$ .

Pull law (iii) inside the square root, then law (iv):

\sqrt{\frac{e^{x}}{e^{7 x}}} \cdot e^{2 x + 5} = \sqrt{e^{-6 x}} \cdot e^{2 x} \cdot e^{5} = e^{-3 x} \cdot e^{2 x} \cdot e^{5} = e^{5} \cdot e^{-x}.

So $A = e^{5} \approx 148.4$ , $k = -1$ .

Remark

Putting an expression in the form $A \, e^{k x}$ before differentiating makes the later derivative step short once the formula for $e^{kx}$ has been established. Whenever a stack of $e$ ‘s appears with sums, products, or roots in the exponent, collapse it first.

Problem 145

Reduce each expression to the form $A \, e^{k x}$ and state $A$ and $k$ .

$e^{3 x} \cdot e^{-x} \cdot e^{2}$ .
$\dfrac{e^{4 x}}{(e^{x})^{3}}$ .
$\bigl(e^{x} \cdot e^{-2 x}\bigr)^{4} \cdot e^{6 x}$ .
$\sqrt[3]{e^{6 x + 9}}$ .

Example 133 (Product and quotient with

e^{x}

)

Differentiate:

$y = (1 + x^{2}) \, e^{x}$ .
$y = \dfrac{1 + e^{x}}{2 x}$ .
By the product rule with $f(x) = 1 + x^{2}$ and $g(x) = e^{x}$ :

\frac{dy}{dx} = (1 + x^{2}) \cdot e^{x} + e^{x} \cdot 2 x = e^{x} (x^{2} + 2 x + 1) = e^{x} (x + 1)^{2}.

The factored form has a useful consequence. Since $e^{x} > 0$ and $(x + 1)^{2} \geq 0$ for every $x$ , the derivative is non-negative everywhere, and zero only at $x = -1$ . By the first derivative rule, $y$ is non-decreasing on the entire real line. The derivative does not change sign around $x = -1$ , so the horizontal tangent there is neither a relative maximum nor a relative minimum. The product structure made this readable; expanding $e^{x} (x^{2} + 2 x + 1)$ would have hidden it.

By the quotient rule with $f(x) = 1 + e^{x}$ and $g(x) = 2 x$ :

\frac{dy}{dx} = \frac{(2 x) \, e^{x} - (1 + e^{x}) \cdot 2}{(2 x)^{2}} = \frac{2 x \, e^{x} - 2 e^{x} - 2}{4 x^{2}} = \frac{x \, e^{x} - e^{x} - 1}{2 x^{2}}.

The numerator does not factor over the elementary tools from this course; the answer is left in the cleaned-up form.

Problem 146

Differentiate each function. Where the answer factors over the polynomial-times- $e^{x}$ family, factor it.

$y = x^{3} \, e^{x}$ .
$y = (x^{2} - 4 x + 1) \, e^{x}$ .
$y = \dfrac{e^{x}}{x}$ .
$y = \dfrac{x^{2}}{e^{x}}$ (write the denominator as $e^{x}$ first; the next section turns this into a one-line chain-rule problem instead).

$e^{-x}$ via the General Power Rule

The function $e^{-x}$ is an exponential with base $1/e \approx 0.37 < 1$ , hence falls under the decreasing-curve case of Recitation 4’s graph discussion. By law (ii) of Recitation 4,

e^{-x} = \frac{1}{e^{x}} = (e^{x})^{-1}.

The general power rule (Recitation 2 / Lesson 4AM toolkit) with $r = -1$ and inner function $g(x) = e^{x}$ supplies the derivative directly, without yet introducing the chain rule for exponentials.

Example 134 (Derivative of

e^{-x}

from the power rule)

Differentiate $y = e^{-x}$ .

Write $y = (e^{x})^{-1}$ . By the general power rule applied to the inner function $g(x) = e^{x}$ , with $g'(x) = e^{x}$ :

\frac{dy}{dx} = -1 \cdot (e^{x})^{-2} \cdot e^{x} = - e^{-2 x} \cdot e^{x} = - e^{-x}.

The derivative of $e^{-x}$ is $-e^{-x}$ . The function reproduces itself under differentiation up to a sign. Geometrically, the slope at every point on $y = e^{-x}$ is the negative of the height. The curve falls everywhere, and its rate of fall scales with how far above the axis it currently sits.

$Graphs of y = e^x (dotted, blue) and y = e^{-x} (solid, red) on the same axes. Both pass through (0, 1). The two curves are mirror images of each other across the y-axis.$

The solid red curve $y = e^{-x}$ is the reflection of $y = e^{x}$ across the $y$ -axis, exactly as predicted by law (ii). Both pass through $(0, 1)$ with slopes $-1$ and $+1$ respectively, mirroring each other.

Problem 147

Differentiate each function by the same trick: rewrite as a power of $e^{x}$ and apply the general power rule.

$y = e^{-2 x}$ .
$y = e^{3 x}$ (write as $(e^{x})^{3}$ ).
$y = \dfrac{1}{e^{2 x}}$ .
Compare answers (1) and (3); explain in one sentence why they are equal.

The general-power-rule manoeuvre runs out as soon as the exponent is anything more interesting than a constant times $x$ . For $e^{x^{2} + 1}$ or $e^{3 x^{2} - 1/x}$ , no choice of $r$ recovers the function. The chain rule is needed.

The Chain Rule for Exponentials

The composition $e^{g(x)}$ is the outer function $f(u) = e^{u}$ applied to the inner function $g(x)$ . The chain rule applied to this composition uses the fact $f' = f$ :

\frac{d}{dx} \bigl[e^{g(x)}\bigr] = f'(g(x)) \cdot g'(x) = e^{g(x)} \cdot g'(x). \tag{8}

Theorem 18 (Chain rule for exponentials)

For every differentiable function $g$ ,

\frac{d}{dx} \bigl[e^{g(x)}\bigr] = e^{g(x)} \cdot g'(x).

In Leibniz form with $u = g(x)$ :

\frac{d}{dx} (e^{u}) = e^{u} \, \frac{du}{dx}. \tag{8a}

The rule says: leave the exponential expression in place, then multiply by the derivative of the inside. The exponential is the only outer function in this course that survives differentiation unchanged; for every other outer function, $f'$ must be looked up separately.

Example 135 (Chain rule applied to

e^{g(x)}

)

Differentiate:

$y = e^{x^{2} + 1}$ .
$y = e^{3 x^{2} - 1/x}$ .
$y = e^{5 x}$ .
$g(x) = x^{2} + 1$ , $g'(x) = 2 x$ . So

\frac{dy}{dx} = e^{x^{2} + 1} \cdot 2 x = 2 x \, e^{x^{2} + 1}.

$g(x) = 3 x^{2} - 1/x = 3 x^{2} - x^{-1}$ , $g'(x) = 6 x + x^{-2} = 6 x + 1/x^{2}$ . So

\frac{dy}{dx} = e^{3 x^{2} - 1/x} \cdot \!\left(6 x + \frac{1}{x^{2}}\right).

$g(x) = 5 x$ , $g'(x) = 5$ . So

\frac{dy}{dx} = e^{5 x} \cdot 5 = 5 \, e^{5 x}.

Part (c) generalises in one sentence: differentiating $e^{k x}$ for any constant $k$ pulls a factor of $k$ out front. This is the formula that powers every applied use in this lesson.

Theorem 19 (Derivative of

e^{k x}

)

For every constant $k$ ,

\frac{d}{dx} (e^{k x}) = k \, e^{k x}.

Example 136 (Differentiating

C e^{k x}

)

Differentiate:

$y = 3 \, e^{5 x}$ .
$y = 3 \, e^{k x}$ , with $k$ constant.
$y = C \, e^{k x}$ , with $C, k$ constants.

By the constant-multiple rule and the formula for $e^{kx}$ :

$\dfrac{dy}{dx} = 3 \cdot 5 \, e^{5 x} = 15 \, e^{5 x}$ .
$\dfrac{dy}{dx} = 3 \cdot k \, e^{k x} = 3 k \, e^{k x}$ .
$\dfrac{dy}{dx} = C \cdot k \, e^{k x} = C k \, e^{k x}$ .

Part (c) earns its own line because, writing $y = C \, e^{k x}$ ,

y' = C k \, e^{k x} = k \cdot (C \, e^{k x}) = k \, y. \tag{9}

The function $y = C \, e^{k x}$ satisfies the rate equation (its own derivative is $k$ times itself) that opened the exponential section of Recitation 4.

Problem 148

Differentiate each function. Each is a one-line application of the formula for $e^{kx}$ or its chain-rule generalisation.

$y = e^{4 x - 7}$ .
$y = 5 \, e^{-x/2}$ .
$y = e^{(x^{2} - 1)^{3}}$ .
$y = \dfrac{e^{2 x}}{x^{2} + 1}$ .
$y = (x^{2} + 1) \, e^{-3 x}$ . Factor the answer; identify every $x$ at which $y' = 0$ .

Problem 149

A drug’s bloodstream concentration after $t$ hours is modelled by

C(t) = 12 \, t \, e^{-t/2} \text{ mg/L}, \qquad t \geq 0.

Use the product rule and the formula for $e^{kx}$ to find $C'(t)$ , factoring fully.
Find the time at which the concentration peaks, and compute the peak value.
By the first derivative test, classify the critical number you found in part (2). State on which intervals the drug is being absorbed faster than it is being eliminated, and on which it is being eliminated faster than it is being absorbed.

The Equation $y' = k \, y$

Equation $(9)$ produced solutions of the proportionality $y' = k \, y$ from formulas of the form $C \, e^{k x}$ . The converse, that every solution looks like this, is the content of the theorem below.

Definition 37 (Differential equation)

A differential equation is an equation expressing a relationship between an unknown function $y$ and one or more of its derivatives. A solution is any function $y = f(x)$ that satisfies the equation when substituted in.

Theorem 20 (Solutions of

y' = k \, y

)

Let $k$ be a constant. The differentiable functions $y = f(x)$ on the real line satisfying

y' = k \, y

are exactly the functions of the form

y = C \, e^{k x},

where $C$ is an arbitrary constant.

Specifying any single value $f(x_{0}) = y_{0}$ determines $C$ uniquely.

Proof

One direction. If $y = C \, e^{k x}$ , then by the formula for $e^{kx}$ and the constant-multiple rule, $y' = C k \, e^{k x} = k \, (C \, e^{k x}) = k \, y$ . The substitution checks.

Other direction. Suppose $y = f(x)$ is differentiable and satisfies $y' = k \, y$ . Define a new function

\varphi(x) = \frac{f(x)}{e^{k x}} = f(x) \, e^{-k x}.

By the product rule applied to $f(x)$ and $e^{-k x}$ , with $\dfrac{d}{dx}(e^{-k x}) = -k \, e^{-k x}$ :

\varphi'(x) = f'(x) \, e^{-k x} + f(x) \cdot (-k) \, e^{-k x} = e^{-k x} \bigl[f'(x) - k \, f(x)\bigr].

The hypothesis $f'(x) = k \, f(x)$ makes the bracket zero. Hence $\varphi'(x) = 0$ for every $x$ .

A standard consequence of the derivative theory is that a differentiable function whose derivative is zero at every point of an interval is constant on that interval. Applying that fact on the real line, $\varphi$ is constant. Call the constant $C$ . Then $\varphi(x) = C$ for every $x$ , and unwinding the definition of $\varphi$ gives

\frac{f(x)}{e^{k x}} = C, \qquad \text{so} \qquad f(x) = C \, e^{k x}.

The single initial condition $f(x_{0}) = y_{0}$ forces $C \, e^{k x_{0}} = y_{0}$ , hence $C = y_{0} \, e^{-k x_{0}}$ , a unique value.

■

The proof’s main move is the substitution $\varphi(x) = f(x) e^{-k x}$ , which extracts the exponential factor through a single derivative computation. The same idea, divide the unknown function by a candidate solution and check that the quotient is constant, reappears in later differential-equation problems.

Example 137 (A pure rate equation)

Find every function $y = f(x)$ such that $y' = -2 y$ .

The equation has the form $y' = k \, y$ with $k = -2$ . By the solutions theorem above,

y = C \, e^{-2 x}, \qquad C \text{ an arbitrary constant}.

For $C > 0$ these are decreasing exponentials. For $C = 0$ the solution is the zero function, and for $C < 0$ the graph lies below the axis and increases toward $0$ .

Example 138 (An initial-value problem)

Find $y = f(x)$ such that $y' = y / 2$ and $f(0) = 4$ .

The equation is $y' = k y$ with $k = 1/2$ , so $y = C \, e^{x/2}$ . The initial condition gives

4 = f(0) = C \, e^{0} = C,

so $C = 4$ and

f(x) = 4 \, e^{x/2}.

The general family from the solutions theorem has one free parameter $C$ ; one data point fixes it.

Note (Procedure: $y' = k y$ with one initial condition)

Read $k$ off the equation $y' = k y$ .
Write the general solution $y = C \, e^{k x}$ .
Plug in the initial condition $f(x_{0}) = y_{0}$ to solve for $C$ .
Substitute $C$ back to obtain the unique solution.

Recitation 4’s exponential scenarios, the 15% investment, the doubling bacteria culture, and the decaying $^{235}\mathrm{U}$ pile, each fit the solutions theorem with a different value of $k$ .

Example 139 (A growing bacteria culture)

A bacteria culture in a Petri dish grows so that its population $N(t)$ satisfies $N'(t) = 0.4 \, N(t)$ , with $t$ measured in hours. The initial population is $N(0) = 500$ . Find $N(t)$ and express the time at which the population first reaches $2000$ in terms of the doubling time.

By the solutions theorem, $N(t) = C \, e^{0.4 \, t}$ . The initial condition gives $500 = C$ , so

N(t) = 500 \, e^{0.4 \, t}.

The population reaches $2000$ when $500 \, e^{0.4 \, t} = 2000$ , that is, $e^{0.4 \, t} = 4$ . Without logarithms, the equation can be reduced as follows. The population-doubling time $\tau$ satisfies $e^{0.4 \, \tau} = 2$ , so $e^{0.4 \, t} = 4 = 2^{2} = (e^{0.4 \, \tau})^{2} = e^{0.8 \, \tau}$ . By the injectivity of $e^{x}$ from Recitation 4, $0.4 \, t = 0.8 \, \tau$ , so $t = 2 \tau$ .

The numeric value of $\tau$ requires a tabulation of $e^{0.4 \, t}$ until the entry $2$ first appears. The structural answer is that the population first reaches four times its initial value at exactly twice its doubling time.

Problem 150

A radioactive sample has mass $M(t)$ in grams at time $t$ years, satisfying $M'(t) = -0.05 \, M(t)$ . Initially $M(0) = 80$ grams.

Write $M(t)$ explicitly using the solutions theorem.
Express the half-life $T$ , the time at which $M(T) = 40$ , by the equation $e^{-0.05 \, T} = 1/2$ , and explain why no further elementary simplification is possible with the tools introduced so far.
Use the injectivity of $e^{x}$ to compute, in terms of $T$ , the time at which only one-eighth of the original sample remains.

Problem 151

An investment of $P$ pounds at continuously compounded annual interest rate $r > 0$ has balance $B(t)$ satisfying $B'(t) = r \, B(t)$ with $B(0) = P$ .

Write $B(t)$ .
Compare two investments at the same time $t > 0$ , one at rate $2 r$ and one at rate $r$ . Show that the ratio of the two balances is $e^{r t}$ , while the ratio of their time-rates of growth is $2 e^{r t}$ . Conclude that the faster investment is growing more than twice as fast for every $t > 0$ , and that its balance is more than twice as large once $t$ exceeds the unique time $T$ for which $e^{r T} = 2$ . Leave $T$ as the solution of $e^{r T} = 2$ .
Two investors start with the same principal $P$ at the same time. One earns rate $r$ , the other earns rate $r/2$ . Show that the first investor’s balance at time $t$ equals the square of the second investor’s balance divided by $P$ for every $t > 0$ . Set up the two balances, reduce both sides to a single $e^{r t}$ by laws of exponents, and conclude by inspection.

Functions $f(x) = e^{k x}$

The formula for $e^{kx}$ plus the solutions theorem set up the family $y = C \, e^{k x}$ as the standard language of proportional rates of change. Two qualitative regimes appear depending on the sign of $k$ .

$Two side panels. Left: y = e^{kx} for k = 0.4, 0.8, 1.2, 1.6, all increasing curves through (0, 1), with steeper k giving faster growth. Right: y = e^{kx} for k = -0.4, -0.8, -1.2, -1.6, all decreasing curves through (0, 1), with more negative k giving faster decay.$

For $k > 0$ (left panel), every curve $y = e^{k x}$ shares four features:

The graph passes through $(0, 1)$ , since $e^{0} = 1$ .
The graph lies strictly above the $x$ -axis: $e^{k x} > 0$ for every real $x$ and every real $k$ .
The $x$ -axis is a horizontal asymptote as $x \to -\infty$ , as seen from the graph of exponential decay in Recitation 4.
The graph is increasing and concave up. The formula for $e^{kx}$ gives $y' = k \, e^{k x} > 0$ and $y'' = k^{2} \, e^{k x} > 0$ .

For $k < 0$ (right panel), the same four facts hold with one reversal:

The graph passes through $(0, 1)$ .
The graph lies strictly above the $x$ -axis.
The $x$ -axis is a horizontal asymptote as $x \to +\infty$ , as seen from the graph of exponential decay in Recitation 4.
The graph is decreasing and concave up. Now $y' = k \, e^{k x} < 0$ but $y'' = k^{2} \, e^{k x} > 0$ .

The graph is concave up in both regimes because $y'' = k^{2} \, e^{k x}$ depends on $k$ only through $k^{2}$ . The exponential family is concave up for every nonzero $k$ .

Problem 152

For each function below, identify $k$ and state whether the curve is increasing or decreasing. Confirm that $y'' > 0$ everywhere by computing $y''$ explicitly.

$y = e^{0.3 \, x}$ .
$y = e^{-1.7 \, x}$ .
$y = 5 \, e^{2 x}$ .
$y = 100 \, e^{-x/4}$ .

Every $b^{x}$ Is an $e^{k x}$

The opening computation reduced $\dfrac{d}{dx}(b^{x})$ to $m_{b} \cdot b^{x}$ with $m_{b}$ an undetermined limit. The formula for $e^{kx}$ reduces $\dfrac{d}{dx}(e^{k x})$ to $k \, e^{k x}$ with $k$ free. The two formulas agree if and only if $b^{x}$ can be rewritten as $e^{k x}$ for the right $k$ .

The graph of $y = e^{x}$ is strictly increasing, stays positive, approaches $0$ on the left, and grows without bound on the right. We use the standard exponential fact, visible in that graph, that $e^{x}$ takes every positive value exactly once. So for every base $b > 0$ there is a unique real number $k$ with

e^{k} = b. \tag{10}

Once $k$ is named, law (iv) of exponents from Recitation 4 supplies the rewrite:

b^{x} = (e^{k})^{x} = e^{k x}. \tag{11}

Theorem 21 (Reduction of

b^{x}

e^{k x}

)

For every base $b > 0$ , there is a unique constant $k$ such that

b^{x} = e^{k x} \qquad \text{for every real } x.

The constant $k$ is the unique solution of $e^{k} = b$ .

Consequently,

\frac{d}{dx} (b^{x}) = k \, b^{x},

where $k$ is the same constant.

The undetermined constant $m_{b}$ from $(6)$ is forced to coincide with $k$ . The numerical experiments at the start of the lesson, $m_{2} \approx 0.693$ , $m_{3} \approx 1.099$ , $m_{5} \approx 1.609$ , are exactly the values of $k$ for which $e^{k} = 2, 3, 5$ respectively. Without logarithms, this lesson cannot compute $k$ algebraically; the existence of $k$ , however, is enough to bring every $b^{x}$ inside the chain-rule machinery.

$The curve y = 2^x drawn solid in red, with y = e^{kx} drawn as a dotted blue overlay, where k is the unique constant satisfying e^k = 2. The two curves coincide everywhere: same shape, two formulas.$

The two formulas $y = 2^{x}$ and $y = e^{k x}$ , with $e^{k} = 2$ , draw the same curve. The choice between them is one of bookkeeping: $e^{k x}$ slots into the chain rule and the differential equation $y' = k y$ without any leftover constants; $b^{x}$ stays close to the original problem statement.

Example 140 (Differentiating

b^{x}

via reduction)

Differentiate $y = 2^{x}$ using the reduction theorem, and compare with the result of $(5)$ .

By the reduction theorem, $2^{x} = e^{k x}$ where $k$ is the unique constant with $e^{k} = 2$ . By the formula for $e^{kx}$ ,

\frac{d}{dx} (2^{x}) = \frac{d}{dx} (e^{k x}) = k \, e^{k x} = k \cdot 2^{x}.

Comparing with equation $(5)$ , $\dfrac{d}{dx}(2^{x}) = m \cdot 2^{x}$ , the constant $m$ from the limit-table calculation must equal the $k$ from the rewrite: $m = k$ , where $e^{k} = 2$ . The two derivations describe the same number through different machinery.

Problem 153

Recitation 4 modelled a British investor’s account balance as $B(t) = 1000 \cdot (1.15)^{t}$ pounds.

Use the reduction theorem to write $B(t) = 1000 \, e^{k t}$ for the unique $k$ with $e^{k} = 1.15$ .
State $B'(t)$ both as $1000 \cdot k \cdot e^{k t}$ and as $k \cdot B(t)$ . Identify the proportionality constant in the rate equation $B'(t) = k \, B(t)$ .
Without computing $k$ numerically, explain why $0 < k < 1$ : since $e^{0} = 1 < 1.15 < e = e^{1}$ and $e^{x}$ is increasing, the unique input $k$ with $e^{k} = 1.15$ must lie between $0$ and $1$ .

Problem 154

A bacteria culture grows according to $N(t) = N_{0} \cdot 3^{t}$ , so the population triples each unit of time.

Use the reduction theorem to rewrite $N(t)$ as $N_{0} \, e^{k t}$ for some $k > 0$ .
Write the rate equation $N'(t) = k \, N(t)$ that $N(t)$ satisfies, with $k$ left as the unique constant satisfying $e^{k} = 3$ .
Numerically, $k \approx 1.099$ from the slope-at-zero table earlier in this lesson. Compute $N'(0)$ and $N'(1)$ when $N_{0} = 100$ .

Problem 155

Let $b > 0$ , $b \neq 1$ , and let $k$ be the unique constant with $e^{k} = b$ . Show that the function $y = (1 + k x) \, b^{x}$ has a single critical number, and locate it. Use the first derivative test to classify the critical point. Then prove that, at that critical $x$ , the height of $y = (1 + k x) \, b^{x}$ is the negative of the height of $y = b^{x}$ .

Note (Toolkit additions from this lesson)

Rule	Form
Derivative of $e^{x}$	$\frac{d}{dx} (e^{x}) = e^{x}$
Chain rule for $e^{g(x)}$	$\frac{d}{dx} \, e^{g(x)} = e^{g(x)} \, g'(x)$
Derivative of $e^{k x}$	$\frac{d}{dx} (e^{k x}) = k \, e^{k x}$
Solutions of $y' = k y$	$y = C \, e^{k x}$ , $C$ arbitrary
Reduction of $b^{x}$	$b^{x} = e^{k x}$ where $e^{k} = b$

Video

Lesson assets

The Exponential Function exe^{x}ex

The Slope of 2x2^{x}2x at x=0x = 0x=0

The Derivative of 2x2^{x}2x at Every xxx

The Number eee

exe^{x}ex in the Toolkit

e−xe^{-x}e−x via the General Power Rule

The Chain Rule for Exponentials

The Equation y′=k yy' = k \, yy′=ky

Functions f(x)=ekxf(x) = e^{k x}f(x)=ekx

Every bxb^{x}bx Is an ekxe^{k x}ekx

The Exponential Function $e^{x}$

The Slope of $2^{x}$ at $x = 0$

The Derivative of $2^{x}$ at Every $x$

The Number $e$

$e^{x}$ in the Toolkit

$e^{-x}$ via the General Power Rule

The Equation $y' = k \, y$

Functions $f(x) = e^{k x}$

Every $b^{x}$ Is an $e^{k x}$