Induction for Derivative Patterns
You know how to differentiate $xe^x$ once. But what does the 50th derivative look like? Induction can prove a general formula for the $n$th derivative of many functions — bridging calculus and proof with a single powerful technique.
Differentiate $xe^x$ once, then again. Without using any formula — what pattern do you notice in the result? Write your prediction for the $n$th derivative below.
Mathematical induction is not limited to algebra and number theory. It is also a powerful tool in calculus for proving patterns in repeated differentiation.
The approach mirrors standard induction, but the inductive step involves differentiating rather than adding:
- Prove the formula holds for the first derivative (base case, $n=1$).
- Assume the formula holds for the $k$th derivative (inductive hypothesis).
- Differentiate once more to prove it holds for the $(k+1)$th derivative (inductive step).
Key facts
- $\dfrac{d^n}{dx^n}(xe^x) = (x+n)e^x$
- $\dfrac{d^n}{dx^n}\!\left(\tfrac{1}{x}\right) = \dfrac{(-1)^n n!}{x^{n+1}}$
- $\dfrac{d^n}{dx^n}(x^n) = n!$
Concepts
- Why the inductive step involves differentiating the assumed $k$th derivative
- How factorial coefficients arise from repeated power-rule differentiation
- How to handle alternating signs via $(-1)^{k+1} = (-1)^k \cdot (-1)$
Skills
- Write a complete induction proof for an $n$th derivative formula
- Correctly apply the product rule and power rule in the inductive step
- Identify and prove new derivative patterns given a conjecture
Statement: Prove by induction that $\dfrac{d^n}{dx^n}(xe^x) = (x + n)e^x$ for all positive integers $n$.
Step 1 — Base Case ($n = 1$):
$$\frac{d}{dx}(xe^x) = e^x + xe^x = (x + 1)e^x$$
This matches the formula with $n = 1$. True for $n = 1$. ✓
Step 2 — Inductive Hypothesis:
Assume $\dfrac{d^k}{dx^k}(xe^x) = (x + k)e^x$ for some positive integer $k$.
Step 3 — Inductive Step ($n = k + 1$):
$$\frac{d^{k+1}}{dx^{k+1}}(xe^x) = \frac{d}{dx}\!\left[(x + k)e^x\right]$$
$$= e^x + (x + k)e^x = (x + k + 1)e^x$$
This is the formula with $n = k + 1$. True for $n = k + 1$.
By the principle of mathematical induction, $\dfrac{d^n}{dx^n}(xe^x) = (x + n)e^x$ for all positive integers $n$. ∎
Statement: Prove by induction that $\dfrac{d^n}{dx^n}(xe^x) = (x + n)e^x$ for all positive integers $n$.
Pause — copy the formula $\frac{d^n}{dx^n}(xe^x)=(x+n)e^x$ and the one-line inductive step (product rule on $(x+k)e^x$) into your book.
Quick check: Using the formula $\dfrac{d^n}{dx^n}(xe^x) = (x+n)e^x$, what is the second derivative of $xe^x$?
We just saw that differentiating $(x+k)e^x$ by the product rule gives $(k+1)e^x+xe^x=(x+(k+1))e^x$, matching the $n=k+1$ formula for $\frac{d^n}{dx^n}(xe^x)$. That raises a question: does differentiating $\frac{(-1)^k k!}{x^{k+1}}$ produce $(-1)^{k+1}$ and $(k+1)!$ naturally for the $1/x$ formula? This card answers it → yes: the chain rule introduces $-(k+1)$ in the numerator, giving $\frac{(-1)^{k+1}(k+1)!}{x^{k+2}}$.
Statement: Prove by induction that $\dfrac{d^n}{dx^n}\!\left(\dfrac{1}{x}\right) = \dfrac{(-1)^n\, n!}{x^{n+1}}$ for all positive integers $n$.
Step 1 — Base Case ($n = 1$):
$$\frac{d}{dx}(x^{-1}) = -x^{-2} = \frac{-1}{x^2}$$
RHS $= \dfrac{(-1)^1 \cdot 1!}{x^{2}} = \dfrac{-1}{x^2}$ ✓
Step 2 — Inductive Hypothesis:
Assume $\dfrac{d^k}{dx^k}\!\left(\dfrac{1}{x}\right) = \dfrac{(-1)^k\, k!}{x^{k+1}}$
Step 3 — Inductive Step ($n = k + 1$):
$$\frac{d^{k+1}}{dx^{k+1}}\!\left(\frac{1}{x}\right) = \frac{d}{dx}\!\left[\frac{(-1)^k\, k!}{x^{k+1}}\right] = (-1)^k k! \cdot \frac{-(k+1)}{x^{k+2}} = \frac{(-1)^{k+1}(k+1)!}{x^{k+2}}$$
This matches the formula with $n = k+1$. By induction, true for all $n \geq 1$. ∎
Statement: Prove by induction that $\dfrac{d^n}{dx^n}\!\left(\dfrac{1}{x}\right) = \dfrac{(-1)^n\, n!}{x^{n+1}}$ for all positive integers $n$.
Pause — copy the formula $\frac{d^n}{dx^n}(1/x)=\frac{(-1)^n n!}{x^{n+1}}$ and the inductive step showing how $(-1)^{k+1}$ and $(k+1)!$ emerge from differentiation into your book.
Did you get this? True or false: The third derivative of $\dfrac{1}{x}$ equals $\dfrac{6}{x^4}$.
Worked examples · 3 in a row, reveal as you go
Compute the first three derivatives of $x^3$ and conjecture the $n$th derivative. Then verify your conjecture for $n = 3$.
Prove by induction that $\dfrac{d^n}{dx^n}(x^n) = n!$ for all positive integers $n$.
Prove by induction that $\dfrac{d^n}{dx^n}(e^{2x}) = 2^n e^{2x}$ for all positive integers $n$. (3 marks)
Fill the gap: Using the formula $\dfrac{d^n}{dx^n}(e^{2x}) = 2^n e^{2x}$, the 4th derivative of $e^{2x}$ is $e^{2x}$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: In the inductive step for $\dfrac{d^n}{dx^n}\!\left(\dfrac{1}{x}\right)$, the factor $(k+1)$ comes from the power rule bringing down the exponent $-(k+1)$.
Activities · practice with the ideas
Use the formula to write down the 5th derivative of $xe^x$.
Compute the 3rd derivative of $\dfrac{1}{x}$ directly (without using the formula), and verify it matches $\dfrac{(-1)^3 \cdot 3!}{x^4}$.
Prove by induction that $\dfrac{d^n}{dx^n}(e^{3x}) = 3^n e^{3x}$ for all positive integers $n$.
State the inductive hypothesis you would use to prove the formula for $\dfrac{d^n}{dx^n}(x^n) = n!$.
What is the 100th derivative of $\dfrac{1}{x}$? Express your answer using factorial notation.
Odd one out: Three of these statements about $n$th derivative induction are correct. Which one is WRONG?
Earlier you predicted what the $n$th derivative of $xe^x$ would look like.
The answer is $\dfrac{d^n}{dx^n}(xe^x) = (x+n)e^x$. Did your prediction come close? The key insight is that each differentiation bumps the constant by $1$ — from $(x+k)e^x$ to $(x+k+1)e^x$ — because differentiating $(x+k)e^x$ by the product rule gives exactly $e^x + (x+k)e^x = (x+k+1)e^x$.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Prove by induction that $\dfrac{d^n}{dx^n}(xe^x) = (x + n)e^x$ for all positive integers $n$. (2 marks)
Q2. Prove by induction that $\dfrac{d^n}{dx^n}\!\left(\dfrac{1}{x}\right) = \dfrac{(-1)^n\, n!}{x^{n+1}}$ for all positive integers $n$. (3 marks)
Q3. Prove by induction that $\dfrac{d^n}{dx^n}(x^n) = n!$ for all positive integers $n$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $n=5$: $(x+5)e^x$ · 2. Direct: $\frac{d}{dx}(x^{-1}) = -x^{-2}$, then $2x^{-3}$, then $-6x^{-4} = \frac{-6}{x^4}$. Formula: $\frac{(-1)^3 \cdot 3!}{x^4} = \frac{-6}{x^4}$ ✓ · 5. $\frac{(-1)^{100} \cdot 100!}{x^{101}} = \frac{100!}{x^{101}}$
Q1 (2 marks): Base case $n=1$: $\frac{d}{dx}(xe^x) = e^x + xe^x = (x+1)e^x = (x+1)e^x$ [1]. Hypothesis: assume $\frac{d^k}{dx^k}(xe^x) = (x+k)e^x$. Step: $\frac{d}{dx}[(x+k)e^x] = e^x + (x+k)e^x = (x+k+1)e^x$ [1]. Conclusion: true for all $n \geq 1$ by induction.
Q2 (3 marks): Base case [1]: $\frac{d}{dx}(x^{-1}) = -x^{-2}$; formula gives $\frac{(-1)^1 1!}{x^2} = \frac{-1}{x^2}$ ✓. Hypothesis: assume $\frac{d^k}{dx^k}(1/x) = \frac{(-1)^k k!}{x^{k+1}}$. Inductive step [1]: $\frac{d}{dx}\!\left[\frac{(-1)^k k!}{x^{k+1}}\right] = (-1)^k k! \cdot \frac{-(k+1)}{x^{k+2}} = \frac{(-1)^{k+1}(k+1)!}{x^{k+2}}$ [1]. This matches the formula for $n=k+1$.
Q3 (3 marks): Base case [1]: $\frac{d}{dx}(x^1) = 1 = 1!$ ✓. Hypothesis: assume $\frac{d^k}{dx^k}(x^k) = k!$. Inductive step [1]: $\frac{d^{k+1}}{dx^{k+1}}(x^{k+1}) = \frac{d^k}{dx^k}[(k+1)x^k] = (k+1) \cdot k! = (k+1)!$ [1]. True for all $n \geq 1$.
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