Differential calculus and mean value theorems notes — Unit 2

In school, you found the first derivative ($dy/dx$ or $y_1$) to find the slope of a tangent. You found the second derivative ($y_2$) to check for maxima or minima. But what if we need to differentiate a function 'n' times?

⇒The Need for the nth Derivative

In engineering, we often approximate complex sine waves or exponential curves into simple polynomials (using Taylor/Maclaurin series). To build an infinite polynomial, we need a generalized formula for the $n$-th derivative ($y_n$).

We don't memorize the nth derivative formulas. We derive them by looking for a pattern after 3 or 4 differentiations. Let's do a 'dry run' for y = (ax+b)ᵐ.

1. 1.Look at $y_3$. Notice the pattern:
2. 2.The coefficient has 3 terms down from m: m(m-1)(m-2)
3. 3.The power of 'a' is 3: a³
4. 4.The exponent is reduced by 3: m-3

The generalized formula $y_n = m(m-1)dots(m-n+1) a^n (ax+b)^{m-n}$ behaves differently depending on the relationship between m (the original power) and n (how many times we differentiate).

The pattern matching trick also works for exponentials and trigonometric functions.

You will rarely be asked to directly find $y_n$ of $1/(x-2)$. Examiners will give you a complex rational function like $y = \frac{x}{x^2 - 3x + 2}$. You MUST break this down using partial fractions before you can apply the standard formulas.

1. 1.1. Factorize the denominator: $(x-1)(x-2)$
2. 2.2. Set up the split: $\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$
3. 3.3. Find $A$ and $B$. $A = -1$, $B = 2$.
4. 4.4. Rewrite $y$: $y = \frac{-1}{x-1} + \frac{2}{x-2}$.
5. 5.5. NOW apply the standard rational $y_n$ formula to both parts.

Final Answer:
$$y_n = (-1) \left[ \frac{(-1)^n n! (1)^n}{(x-1)^{n+1}} \right] + 2 \left[ \frac{(-1)^n n! (1)^n}{(x-2)^{n+1}} \right]$$

$$y_n = \frac{(-1)^{n+1} n!}{(x-1)^{n+1}} + \frac{2(-1)^n n!}{(x-2)^{n+1}}$$

What if you need the $n$-th derivative of two functions multiplied together, like $y = x^2 e^{3x}$? The standard product rule $(uv' + u'v)$ is a nightmare to expand $n$ times. Leibnitz gives us a beautiful shortcut.

Notice anything? It's exactly the Binomial Expansion! Instead of raising to powers, we are differentiating. The coefficients (1, $n$, $n(n-1)/2$, etc.) are identical to Pascal's triangle.

The formula is theoretically infinite. But if we choose '$v$' correctly, the series will truncate (stop) after just a few terms, making the math extremely easy.

Find the nth derivative of $y = x^2 e^{3x}$.

1. 1.Write the formula: $y_n = u_n v + n u_{n-1} v_1 + \frac{n(n-1)}{2} u_{n-2} v_2$
2. 2.Substitute the values directly:
3. 3.$y_n = [3^n e^{3x}] (x^2) + n [3^{n-1} e^{3x}] (2x) + \frac{n(n-1)}{2} [3^{n-2} e^{3x}] (2)$
4. 4.The 2 in the third term cancels with the /2.
5. 5.Factor out the common term e^(3x) to clean it up:

The most guaranteed 10-mark question in Unit 2 is proving a differential equation using Leibnitz. It ALWAYS follows a strict 7-step algorithmic dry run.

1. 1.Step 1: Differentiate once.
2. 2.$y_1 = \cos(m \sin^{-1} x) \cdot \frac{m}{\sqrt{1-x^2}}$
3. 3.Step 2: Cross multiply to remove the root from denominator.
4. 4.$\sqrt{1-x^2} y_1 = m \cos(m \sin^{-1} x)$
5. 5.Step 3: Square both sides to eliminate the square root entirely.
6. 6.$(1-x^2) y_1^2 = m^2 \cos^2(m \sin^{-1} x)$
7. 7.Step 4: Convert any trigonometric terms back into 'y' to simplify.
8. 8.$(1-x^2) y_1^2 = m^2 (1 - \sin^2(m \sin^{-1} x))$
9. 9.$(1-x^2) y_1^2 = m^2 (1 - y^2)$

1. 1.Step 5: Differentiate implicitly one more time.
2. 2.$(1-x^2) [2y_1 y_2] + y_1^2 [-2x] = m^2 [-2yy_1]$
3. 3.Step 6: The magic step. Divide the entire equation by $2y_1$.
4. 4.$(1-x^2)y_2 - xy_1 + m^2 y = 0$
5. 5.Step 7: Apply Leibnitz's theorem 'n' times to the entire equation.

Now, simply add all the expanded terms together and group them by $y_{n+2}$, $y_{n+1}$, and $y_n$.

If you drive a car exactly 60 km in exactly 1 hour, your average (mean) speed was 60 km/h. Mean Value Theorems (MVTs) prove that at some specific instant during that trip, your speedometer MUST have read exactly 60 km/h. You cannot average 60 km/h without actually hitting 60 km/h at least once.

Rolle's Theorem establishes that if a continuous curve goes up, it must come back down to return to its starting height. At the very peak of that turn, the tangent is horizontal (slope = 0).

Verify Rolle's Theorem for $f(x) = (x-2)(x-3)^2$ on $[2, 3]$.

1. 1.1. Since f(x) is a polynomial, it is continuous on [2, 3].
2. 2.2. Since f'(x) = 3x² - 16x + 21 exists everywhere, it is differentiable on (2, 3).
3. 3.3. Find f(a) and f(b):
4. 4. $f(2) = (0)(1)^2 = 0$
5. 5. $f(3) = (1)(0)^2 = 0$
6. 6. Since f(2) = f(3) = 0, all conditions are satisfied.

Now find 'c'. Set f'(c) = 0.

Since $c = 2.33$ lies squarely inside the open interval (2, 3), Rolle's Theorem is verified.

Lagrange removed the third condition of Rolle's theorem (that f(a) must equal f(b)). If the curve starts and ends at different heights, we draw a 'Secant line' between the start and end. LMVT says there will be a 'Tangent line' exactly parallel to this secant.

Notice that if $f(b) = f(a)$, the numerator becomes 0, giving f'(c) = 0. This proves Rolle's theorem is just a special case of Lagrange's!

Verify LMVT for $f(x) = \log x$ on $[1, e]$.

1. 1.1. log(x) is continuous for all x > 0, so it is continuous on [1, e].
2. 2.2. f'(x) = 1/x, which exists on (1, e), so it is differentiable.
3. 3.3. Calculate the endpoints:
4. 4. f(b) = f(e) = log(e) = 1
5. 5. f(a) = f(1) = log(1) = 0

Since e is approx 2.718, $c = 1.718$. This value perfectly lies inside the interval (1, 2.718). LMVT verified!

Cauchy introduces a second function, g(x). It states that under the same continuity and differentiability conditions, there exists a point c such that the ratio of their derivatives equals the ratio of their average changes.

There is one extra condition here: g'(x) must NOT equal 0 anywhere in the interval. If g'(x) was 0, the denominator of the left side would become undefined.

How does a calculator know what sin(37°) is? It doesn't draw a triangle. It uses an infinite polynomial series. Taylor's theorem allows us to convert ANY differentiable function into an infinite polynomial by evaluating its derivatives at a specific point 'a'.

If we set the expansion point to exactly zero ($a=0$), it simplifies beautifully. This special case is named after Maclaurin.

Derive the Maclaurin series expansion for $f(x) = e^x$.

1. 1.We need to find the value of the function and its derivatives at x=0.
2. 2.f(x) = e^x => f(0) = e^0 = 1
3. 3.f'(x) = e^x => f'(0) = e^0 = 1
4. 4.f''(x) = e^x => f''(0) = e^0 = 1
5. 5.f'''(x) = e^x => f'''(0) = e^0 = 1

This is why the derivative of $e^x$ is itself! If you differentiate that entire infinite polynomial, all the terms shift left and recreate the exact same polynomial.

⇒Unit 2 Calculus Mastery

• Can you derive the nth derivative for a rational function using partial fractions?
• Do you remember the rule for choosing 'v' in Leibnitz's theorem?
• Can you perform the 7-step algorithmic dry run to prove a differential equation?
• Do you remember the 3 conditions for Rolle's Theorem?
• Can you accurately find the value of 'c' in LMVT without algebraic errors?
• Can you quickly expand standard functions using Maclaurin's formula?