The derivative $\frac{dy}{dx}$ represents the instantaneous rate of change of the dependent variable $y$ with respect to the independent variable $x$. When both variables depend implicitly on a third parameter—typically time $t$—we employ the chain rule to link their relative rates of change:
\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \]
In related rates problems involving geometric structures (spheres, cones, cylinders), you must construct an equation relating the physical quantities, differentiate implicitly with respect to $t$, and isolate the required rate. A recurring trap in JEE Advanced involves tracking water levels in conic reservoirs, where both the liquid surface radius $r$ and height $h$ vary simultaneously, necessitating the use of similar triangles to eliminate a variable before differentiation.
Worked Examples
1
A spherical balloon is inflated at a constant rate of $30\text{ cm}^3/\text{s}$. Find the rate at which its radius is increasing when the radius reaches $5\text{ cm}$.
Show solution
Let $V$ represent the volume and $r$ denote the radius of the sphere. We are given the constant rate of volume expansion $\frac{dV}{dt} = 30\text{ cm}^3/\text{s}$.
Step 1: Set up the geometric volume formula
The volume of a sphere is given by:
\[ V = \frac{4}{3}\pi r^3 \]
Step 2: Differentiate implicitly with respect to time $t$
Applying the chain rule gives:
\[ \frac{dV}{dt} = \frac{4}{3}\pi \cdot \left(3r^2 \frac{dr}{dt}\right) = 4\pi r^2 \frac{dr}{dt} \]
Step 3: Substitute known values and solve for $\frac{dr}{dt}$
Set $r = 5\text{ cm}$ and $\frac{dV}{dt} = 30$:
\[ 30 = 4\pi (5)^2 \frac{dr}{dt} \implies 30 = 100\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{30}{100\pi} = \frac{3}{10\pi}\text{ cm/s} \]
Final Answer: The radius is increasing at a rate of $\frac{3}{10\pi}\text{ cm/s}$.
✎ Self-Check — 5 questions0 / 5
Q1.A stone is dropped into a quiet lake and waves move in circles at a speed of $4\text{ cm/s}$. At the instant when the radius of the circular wave is $10\text{ cm}$, how fast is the enclosed area increasing?
Q2.The volume of a cube is increasing at the rate of $8\text{ cm}^3/\text{s}$. How fast is the surface area increasing when the length of an edge is $12\text{ cm}$?
Q3.Water is leaking out from a conical tank at a constant rate of $1\text{ cm}^3/\text{s}$ through a tiny hole at the vertex at the bottom. When the slanting height of the water level is $4\text{ cm}$, find the rate of decrease of the slanting height if the semi-vertical angle of the cone is $30^\circ$.
Q4.A man $2\text{ m}$ high walks at a uniform speed of $5\text{ km/h}$ away from a lamp-post which is $6\text{ m}$ high. The rate at which the length of his shadow increases is:
Q5.For a parametric particle trace given by $x = t^3 - 3t$ and $y = t^2 + 4t$, the rate of change of $y$ with respect to $x$ at $t = 2$ is:
Tangents, Normals, and Curves IntersectionTopic 2
For any differentiable function $y = f(x)$, the value of the derivative at a specific point $P(x_0, y_0)$ dictates the geometric slopes of its associated linear lines:
Tangent Line: Passes through $P$ with slope $m_1 = f'(x_0)$. The equation of the tangent line is:
\[ y - y_0 = f'(x_0)(x - x_0) \]
Normal Line: A line perpendicular to the tangent at the point of contact $P$. Its slope is $m_2 = -\frac{1}{f'(x_0)}$, provided $f'(x_0) \neq 0$. The equation of the normal line is:
\[ y - y_0 = -\frac{1}{f'(x_0)}(x - x_0) \]
The angle of intersection $\phi$ between two intersecting curves $y = f(x)$ and $y = g(x)$ is defined as the acute angle between their respective tangents at the point of intersection. It is computed via the formula:
\[ \tan\phi = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \]
If $1 + m_1 m_2 = 0 \implies m_1 m_2 = -1$, the curves intersect at an angle of $90^\circ$, meaning they are strictly orthogonal curves.
Worked Examples
1
Find the equations of the tangent and normal to the curve $y = x^3 - 2x + 1$ at the point where $x = 2$.
Show solution
First, determine the coordinate point $y_0$ by substituting $x_0 = 2$ into the curve equation:
\[ y_0 = (2)^3 - 2(2) + 1 = 8 - 4 + 1 = 5 \implies P(2, 5) \]
Next, compute the general derivative to find the instantaneous slope:
\[ f'(x) = 3x^2 - 2 \implies m_1 = f'(2) = 3(2)^2 - 2 = 12 - 2 = 10 \]
Equation of Tangent:
\[ y - 5 = 10(x - 2) \implies y - 5 = 10x - 20 \implies 10x - y - 15 = 0 \]
Equation of Normal:
The normal slope is $m_2 = -\frac{1}{10}$:
\[ y - 5 = -\frac{1}{10}(x - 2) \implies 10y - 50 = -x + 2 \implies x + 10y - 52 = 0 \]
Final Answer: Tangent: $10x - y - 15 = 0$; Normal: $x + 10y - 52 = 0$.
✎ Self-Check — 5 questions0 / 5
Q1.The angle of intersection between the curves $y = x^2$ and $6y = 7 - x^3$ at the point $(1, 1)$ is:
Q2.If the line $y = x$ is tangent to the curve $y = x^2 + k$, then the value of the constant $k$ must be:
Q3.The slope of the normal to the curve $y = 2x^2 + 3\sin x$ at $x = 0$ is:
Q4.The point on the curve $y = x^2$ where the tangent is parallel to the secant line chord joining $(1, 1)$ and $(2, 4)$ is:
Q5.If the curves $y = e^{ax}$ and $y = e^{bx}$ cut orthogonally, then the product relationship $ab$ must satisfy:
2
Module 2
Monotonicity & Optimization Analysis
Monotonicity and Critical PointsTopic 1
Derivatives serve as the primary tool to track whether a function is climbing or falling over an interval:
Increasing Function: A function $f(x)$ is strictly increasing over an interval if $f'(x) > 0$ for all points in that interval.
Decreasing Function: A function $f(x)$ is strictly decreasing over an interval if $f'(x) < 0$ for all points in that interval.
A function is monotonic if it maintains a single directional trend (either entirely non-increasing or non-decreasing) across its domain. Critical Points are the fundamental domain positions where local behavior can shift. A point $x = c$ in the domain of $f(x)$ is a critical point if:
\[ f'(c) = 0 \quad \text{or} \quad f'(c) \text{ is completely undefined} \]
A common mistake in competitive exams is ignoring points where the derivative is undefined (such as $x=0$ for $f(x)=|x|$), which can be valid locations for local extrema.
Worked Examples
1
Determine the intervals of monotonicity for the function $f(x) = 2x^3 - 3x^2 - 12x + 5$.
Show solution
Differentiate the polynomial function with respect to $x$:
\[ f'(x) = 6x^2 - 6x - 12 \]
To find the critical numbers, set $f'(x) = 0$:
\[ 6(x^2 - x - 2) = 0 \implies 6(x-2)(x+1) = 0 \implies x = 2, \quad x = -1 \]
These two points divide the real number line into three intervals: $(-\infty, -1)$, $(-1, 2)$, and $(2, \infty)$. We analyze the sign of $f'(x)$ in each sub-interval using the sign scheme method:
Final Answer: Strictly increasing on $(-\infty, -1] \cup [2, \infty)$ and strictly decreasing on $[-1, 2]$.
✎ Self-Check — 5 questions0 / 5
Q1.The function $f(x) = x^3 - 3x$ is strictly decreasing in the interval:
Q2.The function $f(x) = \frac{\ln x}{x}$ is strictly increasing in the interval:
Q3.Find the total number of critical points for the transcendental function $f(x) = x - 2\sin x$ on the open interval $(0, 2\pi)$.
Q4.Which of the following functions is strictly decreasing on the interval $(0, \pi/2)$?
Q5.For the function $f(x) = \frac{x}{x^2+1}$, the interval of absolute increase is:
Extrema and Geometric OptimizationTopic 2
Local extreme values occur exclusively at critical domain positions and are analyzed using two standard protocols:
First Derivative Test: Observe the sign changes of $f'(x)$ when passing through a critical point $c$ from left to right. If $f'(x)$ changes from positive to negative, $c$ is a local maximum. If $f'(x)$ changes from negative to positive, $c$ is a local minimum.
Second Derivative Test: If $f'(c) = 0$, evaluate the second derivative at that point:
\[ f''(c) > 0 \implies \text{Local Minimum at } c \]
\[ f''(c) < 0 \implies \text{Local Maximum at } c \]
If $f''(c) = 0$, the test is completely inconclusive, requiring you to look for higher-order derivatives or revert to the First Derivative Test.
To find the Global (Absolute) Extrema on a bounded closed interval $[a, b]$, you collect the functional values at all interior critical points alongside the endpoint values $f(a)$ and $f(b)$. The absolute maximum is the largest value in this set, and the absolute minimum is the smallest.
Worked Examples
1
An open tank with a square base and vertical sides is to be constructed from a metal sheet of area $c^2$. Show that the maximum volume of the tank is $\frac{c^3}{6\sqrt{3}}$.
Show solution
Let x represent the length of the square base side and h denote the vertical height of the tank.
Step 1: Write down the constraint equation
The total surface area of an open box with a square base is:
\[ S = x^2 + 4xh = c^2 \implies 4xh = c^2 - x^2 \implies h = \frac{c^2 - x^2}{4x} \]
Step 2: Formulate the objective volume function
\[ V = x^2 h = x^2 \left( \frac{c^2 - x^2}{4x} \right) = \frac{1}{4}(c^2 x - x^3) \]
Step 3: Optimize using derivatives
Differentiate $V$ with respect to $x$ and set the result to 0:
\[ \frac{dV}{dx} = \frac{1}{4}(c^2 - 3x^2) = 0 \implies 3x^2 = c^2 \implies x = \frac{c}{\sqrt{3}} \]
Check the second derivative to verify it is a maximum:
\[ \frac{d^2V}{dx^2} = \frac{1}{4}(-6x) = -\frac{3x}{2} < 0 \quad (\text{Valid Maximum}) \]
Step 4: Compute the maximum volume expression
Substitute $x = \frac{c}{\sqrt{3}}$ back into the volume formula:
\[ V_{\max} = \frac{1}{4}\left(c^2 \left(\frac{c}{\sqrt{3}}\right) - \left(\frac{c}{\sqrt{3}}\right)^3\right) = \frac{1}{4}\left(\frac{c^3}{\sqrt{3}} - \frac{c^3}{3\sqrt{3}}\right) = \frac{1}{4}\left(\frac{2c^3}{3\sqrt{3}}\right) = \frac{c^3}{6\sqrt{3}} \]
Final Answer: Proved. Maximum volume is $\frac{c^3}{6\sqrt{3}}$.
✎ Self-Check — 5 questions0 / 5
Q1.The minimum value of the objective function $f(x) = x^x$ for $x > 0$ occurs at:
Q2.The maximum value of $f(x) = \frac{x+1}{\sqrt{x^2+1}}$ across the entire real domain is:
Q3.Divide a positive real number $20$ into two components such that the product of the square of one number and the cube of the other is maximized. The parts are:
Q4.The perimeter of a rectangle is bounded at $100\text{ m}$. The maximum available area for this layout configuration is:
Q5.Find the point of local minimum for the function $f(x) = 2x^3 - 21x^2 + 60x + 10$.
3
Module 3
Higher Order Behavior & Curve Geometries
Concavity and Inflection PointsTopic 1
While the first derivative tracks direction, the second derivative controls the curvature or shape of the graph:
Concave Upwards: The graph curves up like a cup. This occurs when $f''(x) > 0$ over an interval, meaning the tangent lines lie entirely below the curve.
Concave Downwards: The graph curves down like a cap. This occurs when $f''(x) < 0$ over an interval, meaning the tangent lines lie entirely above the curve.
An Inflection Point is a transition point on the curve where the concavity changes from concave up to concave down, or vice versa. For a point $x = x_0$ to be an inflection point, $f''(x_0) = 0$ (or is undefined) and the sign of $f''(x)$ must change across $x_0$.
Worked Examples
1
Find the points of inflection for the polynomial function $f(x) = 3x^5 - 5x^4$.
Show solution
Compute the first and second derivatives sequentially:
\[ f'(x) = 15x^4 - 20x^3 \]
\[ f''(x) = 60x^3 - 60x^2 = 60x^2(x - 1) \]
To find candidate inflection points, set $f''(x) = 0$:
\[ 60x^2(x - 1) = 0 \implies x = 0 \quad \text{or} \quad x = 1 \]
Now analyze the sign changes of $f''(x)$ across these candidate points:
Across $x = 0$: For $x < 0$, $f''(x) = 60(+)(-) = -$. For $0 < x < 1$, $f''(x) = 60(+)(-) = -$. Since the sign does not change across $x=0$, it is not an inflection point.
Across $x = 1$: For $0 < x < 1$, $f''(x) = -$. For $x > 1$, $f''(x) = 60(+)(+) = +$. Since the concavity flips from downwards to upwards, $x = 1$ is a valid inflection point.
Find the corresponding y-coordinate: $f(1) = 3(1)^5 - 5(1)^4 = -2$.
Final Answer: The inflection point is $(1, -2)$.
✎ Self-Check — 5 questions0 / 5
Q1.The function $f(x) = x e^x$ is concave downwards in the domain interval:
Q2.How many inflection points exist for the curve equation $y = x^4 - 6x^2 + 5$?
Q3.The curve $y = x^3 - 3x^2 + 6x + 5$ is concave upwards everywhere inside which domain boundary?
Q4.At an inflection point $x = c$ for a thrice differentiable curve map $f(x)$, which condition is unconditionally met?
Q5.For the cubic path $y = 2x^3 - 3x^2 - 12x + 6$, the inflection node location is situated at:
Systematic Curve Sketching, Approximations, and ErrorsTopic 2
To draw a reliable sketch of a function $y = f(x)$, combine several analytical criteria into a step-by-step process:
Domain and Range: Determine the allowed inputs and corresponding outputs.
Intercepts: Find the x-intercepts by setting $y = 0$, and the y-intercept by setting $x = 0$.
Symmetry: Check if the function is even ($f(-x) = f(x)$), odd ($f(-x) = -f(x)$), or periodic.
Asymptotes: Find vertical asymptotes where the function blows up, and horizontal asymptotes as $x \to \pm\infty$.
Monotonicity and Concavity: Map out increasing/decreasing intervals and concavity using $f'(x)$ and $f''(x)$ to position local extrema and inflection points accurately.
Derivatives also enable linear approximations for tracking changes or errors over short distances. The value of a function at a nearby point can be estimated using its linear tangent approximation:
\[ f(a+h) \approx f(a) + h \cdot f'(a) \]
If $x$ changes by a small step $\Delta x$, the resulting change in $y$ is modeled using differentials: $\Delta y \approx f'(x)\Delta x$. We measure these changes using three metrics:
\[ \text{Absolute Error} = \Delta y, \quad \text{Relative Error} = \frac{\Delta y}{y}, \quad \text{Percentage Error} = \left(\frac{\Delta y}{y}\right) \times 100 \]
Worked Examples
1
Use differentials to approximate the value of $\sqrt{25.3}$.
Show solution
Define an appropriate functional framework: $f(x) = \sqrt{x}$.
Choose a convenient baseline value $a = 25$ where the output is an exact integer, and set the small step deviation to $h = 0.3$.
The derivative of our function is:
\[ f'(x) = \frac{1}{2\sqrt{x}} \]
Evaluate the function and its derivative at the baseline value $a = 25$:
\[ f(25) = \sqrt{25} = 5, \quad f'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{10} = 0.1 \]
Apply the linear approximation formula:
\[ f(25 + 0.3) \approx f(25) + (0.3) \cdot f'(25) \implies \sqrt{25.3} \approx 5 + (0.3) \cdot (0.1) = 5 + 0.03 = 5.03 \]
Final Answer: $\sqrt{25.3} \approx 5.03$.
✎ Self-Check — 5 questions0 / 5
Q1.If the error in measuring the radius of a sphere is $1\%$, then the resulting percentage error in calculating its volume is:
Q2.Using differential linear modeling, the approximate scalar estimation for $\sqrt[3]{27.27}$ resolves to:
Q3.While measuring a side length $x$ of an equilateral triangle, a structural error of $2\%$ is introduced. The corresponding deviation error in its computed surface area matches:
Q5.Approximate the real metric functional change of $y = x^2 + 3x$ when $x$ transitions marginally from $2$ down to $1.98$:
4
Module 4
Indeterminate Limits & Mean Value Theorems
L'Hôpital's Rule and Indeterminate FormsTopic 1
L'Hôpital's Rule provides a reliable method for evaluating limits that result in indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. It states that if $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ (or $\pm\infty$), then:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
provided the limit on the right exists and $g'(x) \neq 0$ near $a$. Other indeterminate forms like $0 \cdot \infty$ or $\infty - \infty$ must be algebraically rearranged into standard fractions before applying the rule. For exponential indeterminate forms ($1^\infty, 0^0, \infty^0$), use natural logarithms to convert the expression into a product layout first.
Mean Value Theorem Applications and Root AnalysisTopic 2
Rolle's Theorem and Lagrange's Mean Value Theorem (LMVT) are powerful tools for proving analytical inequalities and identifying the exact roots of transcendental equations.
By analyzing the derivative of a function, you can determine how many real roots it has. For instance, if Rolle's Theorem guarantees that between any two real roots of $f(x) = 0$ there must exist at least one root of $f'(x) = 0$, then by contrapositive logic, if $f'(x) = 0$ has zero real roots across an interval, $f(x) = 0$ can have at most one real root in that interval. This method is effective for proving that an equation has exactly one unique real root.
Worked Examples
1
Prove that the equation $x^3 + 3x + d = 0$ has exactly one real root for any real constant value $d$.
Show solution
Define the polynomial function: $f(x) = x^3 + 3x + d$.
Since $f(x)$ is a polynomial, it is continuous and differentiable everywhere across the real line.
Step 1: Verify existence of at least one root using limits
Analyze the behavior of the function at extreme limits:
\[ \lim_{x \to \infty} f(x) = \infty \quad \text{and} \quad \lim_{x \to -\infty} f(x) = -\infty \]
Since the graph transitions from negative infinity to positive infinity, the Intermediate Value Theorem (IVT) guarantees that the continuous path must cross the x-axis at least once, confirming the existence of at least one real root.
Step 2: Prove uniqueness using monotonicity
Compute the first derivative of the function:
\[ f'(x) = 3x^2 + 3 = 3(x^2 + 1) \]
Since $x^2 \ge 0$ for all real numbers, $3(x^2 + 1) \ge 3 > 0$. Because the derivative is strictly positive everywhere ($f'(x) > 0$), the function is strictly increasing across its entire domain. A strictly increasing function can cross the x-axis at most once. Therefore, the equation cannot have multiple roots.
Final Answer: Proved. The equation has exactly one unique real root.
✎ Self-Check — 5 questions0 / 5
Q1.The total number of real roots for the equation $x^5 + x^3 + x - 5 = 0$ is exactly:
Q2.For the function $f(x) = x^3 - 3x + k$, if Rolle's Theorem is satisfied in the interval $[0, \sqrt{3}]$, then the value of the matching parameter $c$ is:
Q3.Using LMVT analysis over a differentiable path mapping, if $f'(x) \le 5$ for all real numbers and $f(2) = 3$, what is the maximum attainable scalar boundary limit for $f(5)$?
Q4.How many distinct real solutions exist for the equation $x^3 - 3x^2 + 6x - 10 = 0$?
Q5.Between any two real roots of the polynomial equation $e^x \cos x = 1$, how many real roots of $e^x \sin x - 1 = 0$ must exist?
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