Derivatives of eˣ and aˣ — proofs: Exercises

Level 1 — Recognition

(Can you pick the right rule and apply it once?)

L1.1 Differentiate $f (x) = e^{x}$ .

L1.2 Differentiate $g (x) = 7^{x}$ .

L1.3 Differentiate $h (x) = e^{x} + x^{5}$ .

Recall Solution — L1

L1.1 The base is $e$ , the exponent is exactly $x$ . This is the "selfie" case: $f^{'} (x) = e^{x} .$

L1.2 Base $a = 7$ (a fixed number), exponent $x$ (the variable). Use $\frac{d}{d x} a^{x} = a^{x} ln a$ with $a = 7$ : $g^{'} (x) = 7^{x} ln 7.$

L1.3 The derivative of a sum is the sum of derivatives. The first term $e^{x}$ is its own derivative. The second term $x^{5}$ has a fixed exponent and a varying base $x$ , so this is the Power Rule case, $\frac{d}{d x} x^{n} = n x^{n - 1}$ : $h^{'} (x) = e^{x} + 5 x^{4} .$ Notice both rules live in the same line: one term has the exponent moving ( $e^{x}$ ), the other has the base moving ( $x^{5}$ ).

Level 2 — Application

(Now the Chain Rule enters: differentiate the inside too.)

L2.1 Differentiate $y = e^{5 x}$ .

L2.2 Differentiate $y = e^{- 3 x}$ .

L2.3 Differentiate $y = 4 \cdot 2^{x}$ .

L2.4 Differentiate $y = e^{x^{2}}$ .

Recall Solution — L2

The Chain Rule says: if $y = e^{u}$ where $u$ is itself a function of $x$ , then $\frac{d y}{d x} = e^{u} \cdot \frac{d u}{d x}$ . Why this tool? $e^{x}$ is its own derivative only when the exponent is plain $x$ . When the exponent is a bigger expression $u$ , we must pay the extra factor $\frac{d u}{d x}$ — the "inner derivative".

L2.1 $u = 5 x$ , so $\frac{d u}{d x} = 5$ : $y^{'} = e^{5 x} \cdot 5 = 5 e^{5 x} .$

L2.2 $u = - 3 x$ , so $\frac{d u}{d x} = - 3$ : $y^{'} = e^{- 3 x} \cdot (- 3) = - 3 e^{- 3 x} .$ The negative inner derivative makes the slope negative — this is decay.

L2.3 The constant $4$ just rides along (constant-multiple rule). Base $2$ , exponent $x$ : $y^{'} = 4 \cdot 2^{x} ln 2.$

L2.4 $u = x^{2}$ , so $\frac{d u}{d x} = 2 x$ : $y^{'} = e^{x^{2}} \cdot 2 x = 2 x e^{x^{2}} .$

Level 3 — Analysis

(Mixed bases, mixed rules, slopes at points.)

L3.1 Differentiate $y = 2^{x^{2}}$ (base $2$ , general exponent).

L3.2 Find the slope of $y = 3^{x}$ at $x = 0$ . Give an exact and a decimal value.

L3.3 Differentiate $y = x^{2} e^{x}$ (product of a power and an exponential).

L3.4 Differentiate $y = \frac{e ^{x}}{x}$ (quotient).

Recall Solution — L3

L3.1 Convert the base to $e$ so the Chain Rule is clean: $2^{x^{2}} = e^{(l n 2) x^{2}} .$ Here $u = (ln 2) x^{2}$ , so $\frac{d u}{d x} = (ln 2) (2 x) = 2 x ln 2$ . Then $y^{'} = e^{(l n 2) x^{2}} \cdot 2 x ln 2 = 2^{x^{2}} (2 x) ln 2.$

L3.2 $\frac{d}{d x} 3^{x} = 3^{x} ln 3$ . At $x = 0$ we have $3^{0} = 1$ , so the slope is $ln 3 \approx 1.0986.$ This is bigger than $1$ : $3^{x}$ starts off steeper than $e^{x}$ (whose slope-at- $0$ is exactly $1$ ), because $3 > e$ .

L3.3 Product rule $\frac{d}{d x} (uv) = u^{'} v + u v^{'}$ with $u = x^{2}$ ( $u^{'} = 2 x$ ) and $v = e^{x}$ ( $v^{'} = e^{x}$ ): $y^{'} = 2 x e^{x} + x^{2} e^{x} = e^{x} (x^{2} + 2 x) .$

L3.4 Quotient rule $\frac{d}{d x} \frac{u}{v} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ with $u = e^{x}$ , $v = x$ : $y^{'} = \frac{e ^{x} \cdot x - e ^{x} \cdot 1}{x ^{2}} = \frac{e ^{x} ( x - 1 )}{x ^{2}} (x \neq = 0) .$ Note the domain caveat: at $x = 0$ the original function is undefined, so no slope there.

Level 4 — Synthesis

(Combine base-change, chain, product; interpret geometry.)

L4.1 Differentiate $y = 5^{3 x}$ two ways (direct, and via base $e$ ) and check they agree.

L4.2 A tangent line to $y = e^{x}$ is drawn at the point $(a, e^{a})$ . Show the tangent hits the $x$ -axis exactly $1$ unit to the left of $a$ , i.e. at $x = a - 1$ , for every $a$ .

L4.3 For $y = a^{x}$ with $a > 0$ , at what height is the curve climbing at a rate numerically equal to its height? (Find the condition on $a$ .)

Recall Solution — L4

L4.1 Direct: the exponent is $3 x$ , so use $\frac{d}{d x} a^{u} = a^{u} ln a \cdot u^{'}$ with $a = 5$ , $u = 3 x$ , $u^{'} = 3$ : $y^{'} = 5^{3 x} ln 5 \cdot 3 = 3 ln 5 \cdot 5^{3 x} .$ Via base $e$ : $5^{3 x} = e^{(3 l n 5) x}$ , so $u = (3 ln 5) x$ , $u^{'} = 3 ln 5$ : $y^{'} = e^{(3 l n 5) x} \cdot 3 ln 5 = 3 ln 5 \cdot 5^{3 x} . ✓$ Same answer — the two routes must agree because $a = e^{l n a}$ is an identity.

L4.2 (See figure.) At the point $(a, e^{a})$ the slope is $y^{'} (a) = e^{a}$ (self-derivative). The tangent line is $y - e^{a} = e^{a} (x - a) .$ Set $y = 0$ to find where it crosses the $x$ -axis: $- e^{a} = e^{a} (x - a) ⟹ x - a = - 1 ⟹ x = a - 1.$ The $e^{a}$ cancels — that is why the " $1$ unit left" is independent of $a$ . This constant sub-tangent is a signature of $e^{x}$ being its own derivative.

Figure — Derivatives of eˣ and aˣ — proofs

L4.3 "Climbing at a rate equal to its height" means slope $=$ height: $a^{x} ln a = a^{x} .$ Since $a^{x} > 0$ for all $x$ , divide it out: $ln a = 1 ⟹ a = e .$ So only the base $e$ has slope equal to height everywhere — matching the definition of $e$ . For any other $a$ , the slope is $ln a$ times the height (a fixed multiplier, never $1$ unless $a = e$ ). See The number e — definitions.

Level 5 — Mastery

(Prove and generalise — the results themselves.)

L5.1 From first principles, using only $M (a) = lim_{h \to 0} \frac{a ^{h} - 1}{h}$ , show $\frac{d}{d x} a^{x} = M (a) a^{x}$ . Then state what $M (a)$ equals and why.

L5.2 Differentiate the "tower-ish" function $y = x^{x}$ for $x > 0$ . (Hint: it's neither a pure power nor a pure exponential — write it via base $e$ .)

L5.3 Show that $\frac{d ^{n}}{d x ^{n}} e^{k x} = k^{n} e^{k x}$ for every positive integer $n$ (repeated differentiation).

Recall Solution — L5

L5.1 Start from the definition of the derivative applied to $f (x) = a^{x}$ : $f^{'} (x) = lim_{h \to 0} \frac{a ^{x + h} - a ^{x}}{h} .$ Use the exponent law $a^{x + h} = a^{x} a^{h}$ : $= lim_{h \to 0} \frac{a ^{x} a ^{h} - a ^{x}}{h} = lim_{h \to 0} \frac{a ^{x} ( a ^{h} - 1 )}{h} .$ The factor $a^{x}$ does not depend on $h$ , so it slides outside the limit: $= a^{x} lim_{h \to 0} \frac{a ^{h} - 1}{h} = a^{x} M (a) .$ What $M (a)$ is: write $a = e^{l n a}$ , so $a^{x} = e^{(l n a) x}$ ; the chain rule gives $\frac{d}{d x} a^{x} = e^{(l n a) x} \cdot ln a = a^{x} ln a$ . Matching, $M (a) = ln a$ . Geometrically $M (a)$ is the slope of $a^{x}$ at $x = 0$ (where the height is $a^{0} = 1$ ).

L5.2 $x^{x}$ has the variable in both base and exponent, so neither the power rule nor the plain exponential rule fits. Rewrite via base $e$ : $x^{x} = e^{x l n x} .$ Let $u = x ln x$ . By the product rule, $\frac{d u}{d x} = 1 \cdot ln x + x \cdot \frac{1}{x} = ln x + 1$ . Chain rule: $\frac{d}{d x} x^{x} = e^{x l n x} (ln x + 1) = x^{x} (ln x + 1) .$

L5.3 By L2-style chain rule, $\frac{d}{d x} e^{k x} = k e^{k x}$ : each differentiation multiplies by $k$ . Applying it $n$ times pulls out one $k$ each time: $\frac{d}{d x} e^{k x} = k e^{k x}, \frac{d ^{2}}{d x ^{2}} e^{k x} = k^{2} e^{k x}, \dots, \frac{d ^{n}}{d x ^{n}} e^{k x} = k^{n} e^{k x} .$ (Formal induction: base case $n = 1$ shown; if $\frac{d ^{n}}{d x ^{n}} e^{k x} = k^{n} e^{k x}$ , differentiating once more gives $k^{n} \cdot k e^{k x} = k^{n + 1} e^{k x}$ . ✓)

Active recall

Recall Quick self-quiz

$\frac{d}{d x} e^{- 3 x}$ ? ::: $- 3 e^{- 3 x}$ Slope of $3^{x}$ at $x = 0$ ? ::: $ln 3 \approx 1.0986$ $\frac{d}{d x} (x^{2} e^{x})$ ? ::: $e^{x} (x^{2} + 2 x)$ $\frac{d}{d x} \frac{e ^{x}}{x}$ ? ::: $\frac{e ^{x} ( x - 1 )}{x ^{2}}$ Where does the tangent to $e^{x}$ at $(a, e^{a})$ meet the $x$ -axis? ::: at $x = a - 1$ $\frac{d}{d x} x^{x}$ ? ::: $x^{x} (ln x + 1)$ $\frac{d ^{n}}{d x ^{n}} e^{k x}$ ? ::: $k^{n} e^{k x}$

Connections

Derivatives of eˣ and aˣ — proofs — the parent proofs these drills exercise.
Chain Rule — every L2–L5 problem uses it.
Power Rule — the contrast case in L1.3 and the L1/L3/L5 traps.
Natural Logarithm ln x — the constant $ln a$ appearing everywhere.
The number e — definitions — L4.3 recovers " $e$ is the base with slope $=$ height".
Exponential Growth and Decay — the sign of the inner derivative in L2.2.
Limits — first principles definition of derivative — the engine of L5.1.