Apply differential calculus techniques to evaluate derivatives, solve optimization problems, determine limits using L'Hospital's rule, and construct Taylor and Maclaurin series expansions.
Differential calculus studies instantaneous change through the derivative, developed systematically by Isaac Newton and Gottfried Wilhelm Leibniz in the 17th century.
For a function \( y = f(x) \), the derivative measures the limiting ratio of the change in \( y \) to the corresponding change in \( x \).
| Interpretation | Meaning |
|---|---|
| Geometrical | \( f'(a) \) = slope of the tangent to \( y = f(x) \) at \( (a, f(a)) \) |
| Physical | If \( s(t) \) = displacement, then \( \dfrac{ds}{dt} \) = velocity and \( \dfrac{d^2s}{dt^2} \) = acceleration |
| Rate of Change | Instantaneous rate of change of \( y \) with respect to \( x \) |
| Economic | Marginal cost, marginal revenue, marginal profit |
A function is differentiable at \( x = a \) only when:
Reason: Left derivative = \(-1\), Right derivative = \(+1\), they are not equal.
| Function | Derivative | Function | Derivative |
|---|---|---|---|
| \( c \) (constant) | \( 0 \) | \( x^n \) | \( nx^{n-1} \) |
| \( e^x \) | \( e^x \) | \( a^x \) | \( a^x \ln a \) |
| \( \ln x \) | \( \dfrac{1}{x} \) | \( \log_a x \) | \( \dfrac{1}{x \ln a} \) |
| \( \sqrt{x} \) | \( \dfrac{1}{2\sqrt{x}} \) | \( \dfrac{1}{x} \) | \( -\dfrac{1}{x^2} \) |
| Function | Derivative | Function | Derivative |
|---|---|---|---|
| \( \sin x \) | \( \cos x \) | \( \cos x \) | \( -\sin x \) |
| \( \tan x \) | \( \sec^2 x \) | \( \cot x \) | \( -\csc^2 x \) |
| \( \sec x \) | \( \sec x \tan x \) | \( \csc x \) | \( -\csc x \cot x \) |
| Function | Derivative |
|---|---|
| \( \sin^{-1}x \) | \( \dfrac{1}{\sqrt{1-x^2}} \) |
| \( \cos^{-1}x \) | \( -\dfrac{1}{\sqrt{1-x^2}} \) |
| \( \tan^{-1}x \) | \( \dfrac{1}{1+x^2} \) |
| \( \cot^{-1}x \) | \( -\dfrac{1}{1+x^2} \) |
| \( \sec^{-1}x \) | \( \dfrac{1}{|x|\sqrt{x^2-1}} \) |
| \( \csc^{-1}x \) | \( -\dfrac{1}{|x|\sqrt{x^2-1}} \) |
| Function | Derivative | Function | Derivative |
|---|---|---|---|
| \( \sinh x \) | \( \cosh x \) | \( \cosh x \) | \( \sinh x \) |
| \( \tanh x \) | \( \text{sech}^2 x \) | \( \coth x \) | \( -\text{csch}^2 x \) |
For differentiable functions \( u(x) \) and \( v(x) \):
| Rule | Formula |
|---|---|
| Constant Multiple | \( \dfrac{d}{dx}(ku) = k\dfrac{du}{dx} \) |
| Sum / Difference | \( \dfrac{d}{dx}(u \pm v) = \dfrac{du}{dx} \pm \dfrac{dv}{dx} \) |
| Product Rule | \( \dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx} \) |
| Quotient Rule | \( \dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \) |
| Chain Rule | \( \dfrac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \) |
Problem: Differentiate \( y = x^2 e^x \)
Solution:
\[ \frac{dy}{dx} = 2x \cdot e^x + x^2 \cdot e^x = e^x(x^2 + 2x) \]Problem: Differentiate \( y = \dfrac{\sin x}{x} \)
Solution:
\[ \frac{dy}{dx} = \frac{x \cos x - \sin x \cdot 1}{x^2} = \frac{x\cos x - \sin x}{x^2} \]Problem: Differentiate \( y = \sin(x^3) \)
Solution:
\[ \frac{dy}{dx} = \cos(x^3) \cdot 3x^2 = 3x^2\cos(x^3) \]When \( x \) and \( y \) are expressed through a parameter \( t \):
Problem: If \( x = t^2 \) and \( y = t^3 \), find \( \dfrac{dy}{dx} \) and \( \dfrac{d^2y}{dx^2} \)
Solution:
Step 1: \( \dfrac{dx}{dt} = 2t \), \( \dfrac{dy}{dt} = 3t^2 \)
Step 2: First derivative
\[ \frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2} \]Step 3: Second derivative
\[ \frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left(\dfrac{3t}{2}\right)}{\dfrac{dx}{dt}} = \frac{\dfrac{3}{2}}{2t} = \frac{3}{4t} \]Problem: If \( x = a\cos\theta \), \( y = a\sin\theta \), find \( \dfrac{dy}{dx} \)
Solution:
\[ \frac{dx}{d\theta} = -a\sin\theta, \quad \frac{dy}{d\theta} = a\cos\theta \] \[ \frac{dy}{dx} = \frac{a\cos\theta}{-a\sin\theta} = -\cot\theta \]Implicit differentiation applies when an equation relates \( x \) and \( y \) without isolating \( y \).
Differentiate both sides with respect to \( x \), treating \( y \) as \( y(x) \). By the chain rule:
If \( F(x, y) = 0 \) and \( F_y \neq 0 \):
Where \( F_x \) and \( F_y \) are partial derivatives of \( F \) with respect to \( x \) and \( y \).
Problem: Find \( \dfrac{dy}{dx} \) for \( x^2 + y^2 = 25 \)
Solution:
\[ 2x + 2y\frac{dy}{dx} = 0 \] \[ \frac{dy}{dx} = -\frac{x}{y} \]Thus, the tangent slope depends on both coordinates of the point.
Problem: If \( x^3 + y^3 = 3axy \), find \( \dfrac{dy}{dx} \)
Solution:
\[ 3x^2 + 3y^2\frac{dy}{dx} = 3a\left(y + x\frac{dy}{dx}\right) \] \[ 3x^2 + 3y^2\frac{dy}{dx} = 3ay + 3ax\frac{dy}{dx} \] \[ 3y^2\frac{dy}{dx} - 3ax\frac{dy}{dx} = 3ay - 3x^2 \] \[ \frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax} \]Logarithmic differentiation simplifies:
For positive \( u(x) \) and differentiable \( v(x) \), if \( y = u^v \):
Problem: Differentiate \( y = x^x \) (where \( x > 0 \))
Solution:
\[ \ln y = x \ln x \] \[ \frac{y'}{y} = \ln x + 1 \] \[ y' = x^x(\ln x + 1) \]Problem: Differentiate \( y = x^{\sin x} \)
Solution:
\[ \ln y = \sin x \cdot \ln x \] \[ \frac{y'}{y} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \] \[ y' = x^{\sin x}\left[\cos x \ln x + \frac{\sin x}{x}\right] \]Higher-order derivatives describe repeated rates of change.
Notation: \( f''(x) \), \( f^{(n)}(x) \), or \( \dfrac{d^ny}{dx^n} \)
Where: \( m, n \) are non-negative integers and \( a \) is constant.
If \( n > m \), then the \( n \)th derivative of \( x^m \) is zero.
For position \( s(t) \):
| Derivative | Physical Meaning |
|---|---|
| \( s'(t) \) | Velocity |
| \( s''(t) \) | Acceleration |
| \( s'''(t) \) | Jerk |
| \( s^{(4)}(t) \) | Snap / Jounce |
Problem: If \( y = e^{2x} \), find \( y''' \)
Solution:
\[ y''' = 2^3 e^{2x} = 8e^{2x} \]Problem: Find the \( n \)th derivative of \( y = \sin(3x) \)
Solution:
\[ \frac{d^ny}{dx^n} = 3^n \sin\!\left(3x + \frac{n\pi}{2}\right) \]If \( f \) is:
Then there exists at least one \( c \in (a, b) \) such that:
Problem: Verify Rolle's theorem for \( f(x) = x^2 - 4x + 3 \) on \( [1, 3] \)
Solution:
Step 1: \( f(1) = 1 - 4 + 3 = 0 \), \( f(3) = 9 - 12 + 3 = 0 \) ✓
Step 2: \( f'(x) = 2x - 4 = 0 \)
Step 3: \( x = 2 \), and \( 2 \in (1, 3) \) ✓
Therefore: \( c = 2 \)
Problem: Verify Rolle's theorem for \( f(x) = x^3 - 3x^2 + 2x \) on \( [0, 2] \)
Solution:
Step 1: \( f(0) = 0 \), \( f(2) = 8 - 12 + 4 = 0 \) ✓
Step 2: \( f'(x) = 3x^2 - 6x + 2 = 0 \)
Step 3: \( x = \dfrac{6 \pm \sqrt{36 - 24}}{6} = \dfrac{6 \pm \sqrt{12}}{6} = 1 \pm \dfrac{1}{\sqrt{3}} \)
Both values \( 1 - \dfrac{1}{\sqrt{3}} \approx 0.42 \) and \( 1 + \dfrac{1}{\sqrt{3}} \approx 1.58 \) lie in \( (0, 2) \) ✓
If \( f \) is:
Then there exists some \( c \in (a, b) \) such that:
Geometrical Meaning: The tangent at \( c \) is parallel to the secant joining the endpoints.
If \( f \) and \( g \) are:
Then there exists some \( c \in (a, b) \) such that:
| Theorem | Relationship |
|---|---|
| Rolle's Theorem | Equal-endpoint case of Lagrange's theorem |
| Lagrange's Theorem | Follows from Cauchy's theorem by taking \( g(x) = x \) |
Problem: Verify LMVT for \( f(x) = x^2 \) on \( [1, 3] \)
Solution:
Step 1: Average slope:
\[ \frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4 \]Step 2: \( f'(c) = 2c = 4 \)
Step 3: \( c = 2 \), and \( 2 \in (1, 3) \) ✓
Problem: Verify LMVT for \( f(x) = x^3 - 3x \) on \( [0, 2] \)
Solution:
Average slope \( = \dfrac{f(2) - f(0)}{2 - 0} = \dfrac{(8-6) - 0}{2} = 1 \)
\( f'(c) = 3c^2 - 3 = 1 \Rightarrow 3c^2 = 4 \Rightarrow c = \dfrac{2}{\sqrt{3}} \approx 1.15 \in (0, 2) \) ✓
Taylor's theorem represents a sufficiently differentiable function near a chosen point.
Where \( \xi \) lies between \( a \) and \( x \).
Setting \( a = 0 \) in Taylor's theorem gives Maclaurin's expansion:
Problem: Expand \( e^x \) using Maclaurin's series
Solution:
\( f(x) = e^x \Rightarrow f(0) = 1 \)
\( f'(x) = e^x \Rightarrow f'(0) = 1 \)
\( f''(x) = e^x \Rightarrow f''(0) = 1 \)
All derivatives at 0 equal 1. Therefore:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]Problem: Expand \( \sin x \) using Maclaurin's series
Solution:
\( f(0) = 0 \), \( f'(0) = 1 \), \( f''(0) = 0 \), \( f'''(0) = -1 \), \( f^{(4)}(0) = 0 \), ...
\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]Problem: Expand \( \ln x \) about \( x = 1 \) up to 3 terms
Solution:
\( f(x) = \ln x \), \( f(1) = 0 \)
\( f'(x) = \dfrac{1}{x} \Rightarrow f'(1) = 1 \)
\( f''(x) = -\dfrac{1}{x^2} \Rightarrow f''(1) = -1 \)
\( f'''(x) = \dfrac{2}{x^3} \Rightarrow f'''(1) = 2 \)
\[ \ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots \]An indeterminate form is a limit pattern whose appearance alone does not determine the limit's value.
The form \( \dfrac{0}{0} \) can produce different results:
| Expression | Limit as \( x \to 0 \) |
|---|---|
| \( \dfrac{x}{x} \) | \( 1 \) |
| \( \dfrac{x^2}{x} \) | \( 0 \) |
| \( \dfrac{x}{x^2} \) | Unbounded (\( \infty \)) |
| Form | Transformation |
|---|---|
| \( 0 \cdot \infty \) | Convert into a quotient: \( \dfrac{f}{1/g} \) or \( \dfrac{g}{1/f} \) |
| \( \infty - \infty \) | Combine using common denominator or rationalization |
| \( 0^0, \; 1^\infty, \; \infty^0 \) | Set \( y = f(x)^{g(x)} \), then examine \( \ln y = g(x)\ln f(x) \) |
If \( f \) and \( g \) are differentiable near \( a \), \( g'(x) \neq 0 \), and \( \dfrac{f}{g} \) has form \( \dfrac{0}{0} \) or \( \dfrac{\infty}{\infty} \), then:
Provided the limit on the right exists or is infinite.
Problem: Evaluate \( \displaystyle\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} \)
Solution:
Step 1: Form is \( \dfrac{0}{0} \) ✓
Step 2: Apply L'Hospital's rule:
\[ = \lim_{x \to 0} \frac{e^x - 1}{2x} \]Step 3: Still \( \dfrac{0}{0} \), apply again:
\[ = \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2} \]Problem: Evaluate \( \displaystyle\lim_{x \to \infty} \frac{\ln x}{x} \)
Solution:
\[ \lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0 \]Problem: Evaluate \( \displaystyle\lim_{x \to 0^+} x \ln x \)
Solution:
Rewrite as \( \dfrac{\ln x}{1/x} \) (form \( \dfrac{\infty}{\infty} \)):
\[ = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0 \]Interior candidates occur where:
Provided \( f(c) \) exists.
| Sign Change of \( f' \) at \( c \) | Conclusion |
|---|---|
| Positive → Negative | Local Maximum |
| Negative → Positive | Local Minimum |
| No sign change | No extremum established |
If \( f'(c) = 0 \):
| Condition | Conclusion |
|---|---|
| \( f''(c) > 0 \) | Local Minimum |
| \( f''(c) < 0 \) | Local Maximum |
| \( f''(c) = 0 \) | Inconclusive – use first derivative test |
On a closed interval \( [a, b] \):
Problem: Find the extrema of \( f(x) = x^2 - 6x + 5 \)
Solution:
Step 1: \( f'(x) = 2x - 6 = 0 \implies x = 3 \)
Step 2: \( f''(x) = 2 > 0 \implies \) minimum
Step 3: \( f(3) = 9 - 18 + 5 = -4 \)
Conclusion: Absolute minimum value \( -4 \) at \( x = 3 \), with no finite maximum on the real line.
Problem: Find the maximum and minimum values of \( f(x) = 2x^3 - 15x^2 + 36x + 10 \)
Solution:
Step 1: \( f'(x) = 6x^2 - 30x + 36 = 0 \)
\( \Rightarrow x^2 - 5x + 6 = 0 \Rightarrow (x-2)(x-3) = 0 \Rightarrow x = 2, 3 \)
Step 2: \( f''(x) = 12x - 30 \)
\( f''(2) = 24 - 30 = -6 < 0 \Rightarrow \) local maximum at \( x = 2 \)
\( f''(3) = 36 - 30 = 6 > 0 \Rightarrow \) local minimum at \( x = 3 \)
Step 3: \( f(2) = 16 - 60 + 72 + 10 = 38 \) (max)
\( f(3) = 54 - 135 + 108 + 10 = 37 \) (min)
Problem: Find absolute extrema of \( f(x) = x^3 - 3x^2 + 1 \) on \( [-1, 4] \)
Solution:
Step 1: \( f'(x) = 3x^2 - 6x = 3x(x-2) = 0 \Rightarrow x = 0, 2 \)
Step 2: Evaluate:
\( f(-1) = -1 - 3 + 1 = -3 \)
\( f(0) = 1 \)
\( f(2) = 8 - 12 + 1 = -3 \)
\( f(4) = 64 - 48 + 1 = 17 \)
Step 3: Absolute Maximum = \( 17 \) at \( x = 4 \)
Absolute Minimum = \( -3 \) at \( x = -1 \) and \( x = 2 \)
| Topic | Key Formula / Concept |
|---|---|
| Derivative Definition | \( f'(x) = \displaystyle\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \) |
| Product Rule | \( (uv)' = u'v + uv' \) |
| Quotient Rule | \( \left(\dfrac{u}{v}\right)' = \dfrac{vu' - uv'}{v^2} \) |
| Chain Rule | \( \dfrac{d}{dx}[f(g(x))] = f'(g(x))g'(x) \) |
| Parametric 1st Derivative | \( \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} \) |
| Parametric 2nd Derivative | \( \dfrac{d^2y}{dx^2} = \dfrac{d/dt(dy/dx)}{dx/dt} \) |
| Implicit Differentiation | \( \dfrac{dy}{dx} = -\dfrac{F_x}{F_y} \) |
| Logarithmic Differentiation | \( y' = u^v\!\left[v'\ln u + v\dfrac{u'}{u}\right] \) |
| \( n \)th Derivative of \( x^m \) | \( \dfrac{m!}{(m-n)!}x^{m-n} \) |
| \( n \)th Derivative of \( e^{ax} \) | \( a^n e^{ax} \) |
| \( n \)th Derivative of \( \sin ax \) | \( a^n \sin(ax + n\pi/2) \) |
| \( n \)th Derivative of \( \cos ax \) | \( a^n \cos(ax + n\pi/2) \) |
| Rolle's Theorem | \( f'(c) = 0 \) when \( f(a) = f(b) \) |
| Lagrange's MVT | \( f'(c) = \dfrac{f(b)-f(a)}{b-a} \) |
| Cauchy's MVT | \( \dfrac{f'(c)}{g'(c)} = \dfrac{f(b)-f(a)}{g(b)-g(a)} \) |
| Taylor Series | \( f(x) = \sum \dfrac{(x-a)^n}{n!}f^{(n)}(a) \) |
| Maclaurin Series | \( f(x) = \sum \dfrac{x^n}{n!}f^{(n)}(0) \) |
| L'Hospital's Rule | \( \lim \dfrac{f}{g} = \lim \dfrac{f'}{g'} \) for \( \dfrac{0}{0} \) or \( \dfrac{\infty}{\infty} \) |
| Second Derivative Test | \( f''(c) > 0 \): min, \( f''(c) < 0 \): max |
| Function | Maclaurin Series | Valid For |
|---|---|---|
| \( e^x \) | \( 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots \) | All \( x \) |
| \( \sin x \) | \( x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots \) | All \( x \) |
| \( \cos x \) | \( 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots \) | All \( x \) |
| \( \ln(1+x) \) | \( x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots \) | \( |x| < 1 \) |
| \( \tan^{-1}x \) | \( x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots \) | \( |x| \leq 1 \) |
| \( (1+x)^n \) | \( 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \cdots \) | \( |x| < 1 \) |
Find \( \dfrac{dy}{dx} \) if \( x = t + \dfrac{1}{t} \), \( y = t - \dfrac{1}{t} \)
Differentiate \( y = x^{\sin x} \) using logarithmic differentiation
Verify Rolle's theorem for \( f(x) = x^3 - 3x^2 + 2x \) on \( [0, 2] \)
Evaluate \( \displaystyle\lim_{x \to 0} \frac{\tan x - x}{x^3} \)
Find the maximum and minimum values of \( f(x) = 2x^3 - 15x^2 + 36x + 10 \)
Expand \( \ln(1+x) \) in powers of \( x \) using Maclaurin's series
Find the \( n \)th derivative of \( y = \sin(3x) \)
If \( x^3 + y^3 = 3axy \), find \( \dfrac{dy}{dx} \)
\( x = t + \dfrac{1}{t} \), \( y = t - \dfrac{1}{t} \)
\[ \frac{dx}{dt} = 1 - \frac{1}{t^2}, \quad \frac{dy}{dt} = 1 + \frac{1}{t^2} \] \[ \frac{dy}{dx} = \frac{1 + 1/t^2}{1 - 1/t^2} = \frac{t^2 + 1}{t^2 - 1} \]\( y = x^{\sin x} \)
\[ \ln y = \sin x \cdot \ln x \] \[ \frac{y'}{y} = \cos x \ln x + \frac{\sin x}{x} \] \[ y' = x^{\sin x}\left[\cos x \ln x + \frac{\sin x}{x}\right] \]\( f(x) = x^3 - 3x^2 + 2x \) on \( [0, 2] \)
\( f(0) = 0 \), \( f(2) = 8 - 12 + 4 = 0 \) ✓
\( f'(x) = 3x^2 - 6x + 2 = 0 \)
\( x = 1 \pm \dfrac{1}{\sqrt{3}} \approx 0.42, 1.58 \) — both in \( (0, 2) \) ✓
\( \displaystyle\lim_{x \to 0} \frac{\tan x - x}{x^3} \)
Form \( \dfrac{0}{0} \). Apply L'Hospital:
\[ = \lim_{x \to 0} \frac{\sec^2 x - 1}{3x^2} = \lim_{x \to 0} \frac{\tan^2 x}{3x^2} \] \[ = \lim_{x \to 0} \frac{1}{3}\left(\frac{\tan x}{x}\right)^2 = \frac{1}{3} \]\( f(x) = 2x^3 - 15x^2 + 36x + 10 \)
\( f'(x) = 6x^2 - 30x + 36 = 0 \Rightarrow x = 2, 3 \)
\( f''(x) = 12x - 30 \)
\( f''(2) = -6 < 0 \Rightarrow \) max at \( x = 2 \), \( f(2) = 38 \)
\( f''(3) = 6 > 0 \Rightarrow \) min at \( x = 3 \), \( f(3) = 37 \)
\( \ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots \) for \( |x| < 1 \)
\( y = \sin(3x) \)
\[ \frac{d^ny}{dx^n} = 3^n \sin\!\left(3x + \frac{n\pi}{2}\right) \]\( x^3 + y^3 = 3axy \)
\[ 3x^2 + 3y^2\frac{dy}{dx} = 3a\left(y + x\frac{dy}{dx}\right) \] \[ \frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax} \]| Reference | Book |
|---|---|
| T-1 | Advanced Engineering Mathematics – R.K. Jain, S.R.K. Iyengar (Narosa Publishing House) |
| R-1 | Higher Engineering Mathematics – B.S. Grewal (Khanna Publishers) |
| R-2 | Mathematics Textbook for Class XII Part I – NCERT |
| R-3 | Mathematics Textbook for Class XII Part II – NCERT |
| OR-1 | https://ncert.nic.in/textbook.php?lemh1=5-6 |
| OR-2 | https://ncert.nic.in/textbook.php?lemh2=1-7 |
After studying this unit, students should be able to:
| CO2 Components | Covered in Section |
|---|---|
| Evaluate derivatives | II, III, IV, V, VI |
| Solve optimization problems | XII |
| Determine limits using L'Hospital's rule | XI |
| Construct Taylor and Maclaurin series | IX |
⭐ End of Unit II Notes ⭐
Differential Calculus and Its Applications
MTH165 – Mathematics for Engineers
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