Unit 4 - Differential Calculus
Required Prior Knowledge
Questions
Consider \(f\left(x\right)=e^{3x}\)
a) Find \(f’\left(x\right) \)
b) Find \(f’’\left(x\right) \)
c) Find \(f’’’\left(x\right) \)
Can you spot any patterns?
Can you predict the 10th derivative?
Can you conjecture a general expression for the \(n\)th derivative?
Solutions
Get Ready
Questions
What does the first derivative tell us about the function?
What does the second derivative tell us about the function?
What about the third, fourth, fifth, … derivatives?
Solutions
Notes
\(\frac{d}{dx}\) is an operator which means it can be applied to any function.
So \(\frac{d}{dx}\left(y\right)\) means the derivative of \(y\) and \(\frac{d}{dx}\left(\frac{dy}{dx}\right)\) means the derivative of \(\frac{dy}{dx}\).$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d^{2}y}{dx^{2}}$$
We use two notations:$$\begin{align}f’\left(x\right)&=\frac{dy}{dx}\\f’’\left(x\right)&=\frac{d^{2}y}{dx^{2}}\\f’’’\left(x\right)&=\frac{d^{3}y}{dx^{3}}\\f^{\left(4\right)}\left(x\right)&=\frac{d^{4}y}{dx^{4}}\end{align}$$
NOTE on notation$$f^{\left(n\right)}\left(x\right)=\frac{d^{n}y}{dx^{n}}$$ is the \(n\)th derivative of \(f\), whereas $$f^{n}\left(x\right)=f\left(f\left(f\left(…\right)\right)\right)$$is the composition of \(f\) with itself \(n\) times.
Examples and Your Turns
Example
Find the first five derivatives of $$f\left(x\right)=x^{4}-3x^{2}+2x-1$$
Your Turn
A particle moves in a straight line so that its position from a fixed point after \(t\) seconds is given by \(s\left(t\right)=3t+5t^{2}-t^{3}\), with \(s\) in cm.
a) Find the velocity of the particle when \(t=2\).
b) Find the acceleration of the particle when \(t=2\).
c) Find the jerk of the particle when \(t=2\).
Example
If \(y=\frac{1}{1-x}\) find \(\frac{dy}{dx}\). Hence, using the principle of mathematical induction, prove that, for all \(n\in\mathbb{Z}^{+}\)$$\frac{d^{n}y}{dx^{n}}=\frac{n!}{\left(1-x\right)^{n+1}}$$
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Step 1 (\(n=1\))
We compute the first derivative of \(y = (1-x)^{-1}\) using the chain rule: $$\begin{align}\frac{dy}{dx} &= -1(1-x)^{-2}(-1) \\&= \frac{1}{(1-x)^2}\end{align}$$Using the provided formula for \(n=1\), we get $$\begin{align}\frac{d^1y}{dx^1} &= \frac{1!}{(1-x)^{1+1}} \\&= \frac{1}{(1-x)^2}\end{align}$$Since the derivative and the formula match, the base case holds.
Step 2 (\(n=k\))
Assume the statement is true for \(n=k\): $$\frac{d^k y}{dx^k} = \frac{k!}{(1-x)^{k+1}}$$
Step 3 (\(n=k+1\))
WTS $$\frac{d^{k+1} y}{dx^{k+1}} = \frac{(k+1)!}{(1-x)^{k+2}}$$ We find the \((k+1)\)-th derivative by differentiating the \(k\)-th derivative with respect to \(x\): $$\begin{align}LHS = \frac{d}{dx}\left[\frac{d^k y}{dx^k}\right] &= \frac{d}{dx}\left[\frac{k!}{(1-x)^{k+1}}\right] \\&= \frac{d}{dx}\left[k!(1-x)^{-(k+1)}\right] \\&= k! \cdot -(k+1)(1-x)^{-(k+1)-1} \cdot (-1) \\&= k!(k+1)(1-x)^{-(k+2)} \\&= \frac{(k+1)!}{(1-x)^{k+2}}=RHS\end{align}$$This matches the required formula for \(n=k+1\).
Step 4 (Conclusion)
By the principle of mathematical induction, since the statement holds for \(n=1\), and assuming it is true for \(n=k\) guarantees it is true for \(n=k+1\), the formula \(\frac{d^n y}{dx^n} = \frac{n!}{(1-x)^{n+1}}\) is true for all \(n \in \mathbb{Z}^+\).
Your Turn
Consider the function \(f\left(x\right)=ax^{n}\).
a) Find \(f’\left(x\right), f’’\left(x\right), f’’’\left(x\right), f^{\left(4\right)}\left(x\right), f^{\left(5\right)}\left(x\right)\)
b) Hence find an expression for \(f^{\left(m\right)}\left(x\right)\)
c) Prove your result using the method of mathematical induction.
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Step 1 (\(m=1\))
We compute the first derivative of \(f(x) = ax^n\) using the power rule: $$\begin{align}f'(x) &= \frac{d}{dx}(ax^n) \\&= anx^{n-1}\end{align}$$Using the provided formula for \(m=1\), we get $$\begin{align}f^{(1)}(x) &= \frac{an!}{(n-1)!}x^{n-1} \\&= \frac{a \cdot n(n-1)!}{(n-1)!}x^{n-1} \\&= anx^{n-1}\end{align}$$Since the derivative and the formula match, the base case holds.
Step 2 (\(m=k\))
Assume the statement is true for \(m=k\): $$f^{(k)}(x) = \frac{an!}{(n-k)!}x^{n-k}$$
Step 3 (\(m=k+1\))
WTS $$f^{(k+1)}(x) = \frac{an!}{(n-(k+1))!}x^{n-(k+1)}$$ We find the \((k+1)\)-th derivative by differentiating the \(k\)-th derivative with respect to \(x\): $$\begin{align}LHS &= \frac{d}{dx}\left[f^{(k)}(x)\right] \\&= \frac{d}{dx}\left[\frac{an!}{(n-k)!}x^{n-k}\right] \\&= \frac{an!}{(n-k)!} \cdot (n-k)x^{n-k-1} \\&= \frac{an!(n-k)}{(n-k)(n-k-1)!}x^{n-(k+1)} \\&= \frac{an!}{(n-(k+1))!}x^{n-(k+1)}=RHS\end{align}$$This matches the required formula for \(m=k+1\).
Step 4 (Conclusion)
By the principle of mathematical induction, since the statement holds for \(m=1\), and assuming it is true for \(m=k\) guarantees it is true for \(m=k+1\), the formula \(f^{(m)}(x)=\frac{an!}{(n-m)!}x^{n-m}\) is true for all positive integers \(m \leq n\).
Your Turn
Thinking back to your conjecture in the Required Prior Knowledge, use mathematical induction to prove your conjecture.
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Step 1 (\(n=1\))
We compute the first derivative of \(f(x) = e^{3x}\) using the chain rule: $$\begin{align}f'(x) &= \frac{d}{dx}(e^{3x}) \\&= 3e^{3x}\end{align}$$Using the provided formula for \(n=1\), we get $$\begin{align}f^{(1)}(x) &= 3^1e^{3x} \\&= 3e^{3x}\end{align}$$Since the derivative and the formula match, the base case holds.
Step 2 (\(n=k\))
Assume the statement is true for \(n=k\): $$f^{(k)}(x) = 3^ke^{3x}$$
Step 3 (\(n=k+1\))
WTS $$f^{(k+1)}(x) = 3^{k+1}e^{3x}$$ We find the \((k+1)\)-th derivative by differentiating the \(k\)-th derivative with respect to \(x\): $$\begin{align}LHS = \frac{d}{dx}\left[f^{(k)}(x)\right] &= \frac{d}{dx}\left[3^ke^{3x}\right] \\&= 3^k \cdot \frac{d}{dx}\left[e^{3x}\right] \\&= 3^k \cdot 3e^{3x} \\&= 3^{k+1}e^{3x}=RHS\end{align}$$This matches the required formula for \(n=k+1\).
Step 4 (Conclusion)
By the principle of mathematical induction, since the statement holds for \(n=1\), and assuming it is true for \(n=k\) guarantees it is true for \(n=k+1\), the formula \(f^{(n)}(x)=3^ne^{3x}\) is true for all positive integers \(n\).
Your Turn
Use mathematical induction to prove that if \(y=\ln x\) then $$\frac{d^{n}y}{dx^{n}}=\frac{\left(-1\right)^{n-1}\left(n-1\right)!}{x^{n}}$$for \(n\in\mathbb{Z}^{+}\).
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Step 1 (\(n=1\))
We compute the first derivative of \(y = \ln x\): $$\begin{align}\frac{dy}{dx} &= \frac{1}{x} \\&= x^{-1}\end{align}$$Using the provided formula for \(n=1\), we get $$\begin{align}\frac{d^1y}{dx^1} &= \frac{(-1)^{1-1}(1-1)!}{x^1} \\&= \frac{(-1)^0 0!}{x} \\&= \frac{1}{x}\end{align}$$Since the derivative and the formula match, the base case holds.
Step 2 (\(n=k\))
Assume the statement is true for \(n=k\): $$\frac{d^k y}{dx^k} = \frac{(-1)^{k-1}(k-1)!}{x^k}$$
Step 3 (\(n=k+1\))
WTS $$\frac{d^{k+1} y}{dx^{k+1}} = \frac{(-1)^{(k+1)-1}((k+1)-1)!}{x^{k+1}} = \frac{(-1)^k k!}{x^{k+1}}$$ We find the \((k+1)\)-th derivative by differentiating the \(k\)-th derivative with respect to \(x\): $$\begin{align}LHS &= \frac{d}{dx}\left[\frac{d^k y}{dx^k}\right] \\&= \frac{d}{dx}\left[(-1)^{k-1}(k-1)! \cdot x^{-k}\right] \\&= (-1)^{k-1}(k-1)! \cdot (-k)x^{-k-1} \\&= (-1)^{k-1} \cdot -1 \cdot k \cdot (k-1)! \cdot x^{-(k+1)} \\&= (-1)^k k! \cdot x^{-(k+1)} \\&= \frac{(-1)^k k!}{x^{k+1}}=RHS\end{align}$$This matches the required formula for \(n=k+1\).
Step 4 (Conclusion)
By the principle of mathematical induction, since the statement holds for \(n=1\), and assuming it is true for \(n=k\) guarantees it is true for \(n=k+1\), the formula \(\frac{d^n y}{dx^n} = \frac{(-1)^{n-1}(n-1)!}{x^n}\) is true for all \(n \in \mathbb{Z}^+\).
Your Turn
Given that \(f\left(x\right)=xe^{x}\) prove that \(f^{\left(n\right)}\left(x\right)=\left(x+n\right)e^{x}\).
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Step 1 (\(n=1\))
We compute the first derivative of \(f(x)\) using the product rule: $$f’(x) = \frac{d}{dx}(xe^x) = (1)(e^x) + (x)(e^x) = e^x + xe^x = (x+1)e^x$$ Using the provided formula for \(n=1\), we get \(f^{\left(1\right)}\left(x\right)=(x+1)e^x\). Since the derivative and the formula match, the base case holds.
Step 2 (\(n=k\))
Assume the statement is true for \(n=k\): $$f^{(k)}(x) = (x+k)e^x$$
Step 3 (\(n=k+1\))
WTS $$f^{(k+1)}(x) = (x+k+1)e^x$$ We find the \((k+1)\)-th derivative by differentiating the \(k\)-th derivative with respect to \(x\): $$\begin{align}LHS = \frac{d}{dx}\left[f^{(k)}(x)\right] &= \frac{d}{dx}\left[(x+k)e^x\right] \\&= \left[\frac{d}{dx}(x+k)\right]e^x + (x+k)\left[\frac{d}{dx}(e^x)\right] \\&= (1)e^x + (x+k)e^x \\&= e^x(1 + x + k) \\&= (x+k+1)e^x=RHS\end{align}$$ This matches the required formula for \(n=k+1\).
Step 4 (Conclusion)
By the principle of mathematical induction, since the statement holds for \(n=1\), and assuming it is true for \(n=k\) guarantees it is true for \(n=k+1\), the formula \(f^{(n)}(x)=(x+n)e^x\) is true for all positive integers \(n\).
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Finding \(f'''(x)\)
We differentiate \(f''(x) = u''v + 2u'v' + uv''\) using the product rule:
$$\begin{align}f'''(x) &= (u''v)' + (2u'v')' + (uv'')' \\&= (u'''v + u''v') + 2(u''v' + u'v'') + (u'v'' + uv''') \\&= u'''v + u''v' + 2u''v' + 2u'v'' + u'v'' + uv''' \\&= u'''v + 3u''v' + 3u'v'' + uv'''\end{align}$$
Finding \(f^{(4)}(x)\)
We differentiate \(f'''(x)\) using the product rule again:
$$\begin{align}f^{(4)}(x) &= (u'''v)' + 3(u''v')' + 3(u'v'')' + (uv''')' \\&= (u^{(4)}v + u'''v') + 3(u'''v' + u''v'') + 3(u''v'' + u'v''') + (u'v''' + uv^{(4)}) \\&= u^{(4)}v + u'''v' + 3u'''v' + 3u''v'' + 3u''v'' + 3u'v''' + u'v''' + uv^{(4)} \\&= u^{(4)}v + 4u'''v' + 6u''v'' + 4u'v''' + uv^{(4)}\end{align}$$
Finding \(f^{(5)}(x)\)
Following the established pattern, we differentiate \(f^{(4)}(x)\). I have added a line break in the expanded step to prevent it from going off the page: $$\begin{align}f^{(5)}(x) &= (u^{(4)}v)' + 4(u'''v')' + 6(u''v'')' + 4(u'v''')' + (uv^{(4)})' \\&= (u^{(5)}v + u^{(4)}v') + 4(u^{(4)}v' + u'''v'') + 6(u'''v'' + u''v''') \\&\quad + 4(u''v''' + u'v^{(4)}) + (u'v^{(4)} + uv^{(5)}) \\&= u^{(5)}v + 5u^{(4)}v' + 10u'''v'' + 10u''v''' + 5u'v^{(4)} + uv^{(5)}\end{align}$$
Conjecture (Leibniz's Formula)
Notice that the coefficients of the terms in the \(n\)-th derivative exactly match the entries in the \(n\)-th row of Pascal's Triangle (1; 1, 1; 1, 2, 1; 1, 3, 3, 1; 1, 4, 6, 4, 1; etc.). These are the binomial coefficients \(\binom{n}{k}\). Additionally, in each term, the sum of the orders of the derivatives of \(u\) and \(v\) is always \(n\).
Comparing this to the Binomial Theorem expansion for \((a+b)^n\), we can state Leibniz's formula for the \(n\)-th derivative of a product:
$$f^{(n)}(x) = \sum_{k=0}^{n} \binom{n}{k} u^{(n-k)}v^{(k)}$$
(Note: \(u^{(0)}\) is simply \(u\) and \(v^{(0)}\) is \(v\))
Investigation
We already know that if \(f\left(x\right)=uv\) then \(f’\left(x\right)=u’v+uv’\) using the product rule.
We can find \(f’’\left(x\right)\) using the product rule again$$\begin{aligned} f’’(x) &= (u’v)’ + (uv’)’ \\ &= (u’’v + u’v’) + (u’v’ + uv’‘) \\ &= u’’v + 2u’v’ + uv’’ \end{aligned}$$
Using a similar approach, find \(f’’’\left(x\right)\), \(f^{\left(4\right)}\left(x\right)\) and \(f^{\left(5\right)}\left(x\right)\).
Can you make a conjecture? Think about the Binomial Theorem.
The general case is caller Leibniz’s formula.
Key Facts
Use this applet to generate a prompt for a Key Fact that you need to know for the course. The idea is that you should KNOW these key facts in order to be able to solve problems.
Taking it Deeper
Conceptual Questions to Consider
Why do functions involving \(e^{x}\) often lead to a proof by induction question?
Why do factorials often appear in proof by induction questions involving derivatives?
What is the difference between the notations \(f^{n}\left(x\right)\) and \(f^{\left(n\right)}\left(x\right)\)?
Common Mistakes / Misconceptions
The most difficult part of proof by induction questions involving derivatives is normally the factorial. Remember that \(k!=\times \left(k-1\right)!\).
It is also common not to show the base case properly. You must differentiate the function using derivative methods, AND show this is the same as the formula given.
Particularly with those that alternate signs, it can be tricky to factorise out the \(-1\). Remember the way to alternate signs is to use \(\left(-1\right)^{n}\).
Connecting This to Other Skills
You need to know the methods of Proof By Induction (1.8) to access this skill. You also need to know all the derivative rules such as Differentiating Polynomials (4.5), Chain Rule (4.7), Product Rule (4.8), Quotient Rule (4.9), Differentiating Exponentials and Logarithms (4.11) and Differentiating Trig (4.12).
Higher derivatives also form the basics of what we will see in Maclaurin Series (8.2) and Maclaurin Series from Differential Equations (8.9).
Self-Reflection
What was the most challenging part of this skill for you?
What are you still unsure about that you need to review?