# 4.4 Application Of Derivatives ## Detailed Theory: Applications of Derivatives ### **1. Introduction to Applications of Derivatives** #### **1.1 Why Study Applications?** Derivatives are not just abstract mathematical concepts; they have powerful real-world applications in: * Finding maximum/minimum values (optimization) * Analyzing rates of change * Understanding curve behavior * Solving physics and engineering problems #### **1.2 The Derivative as a Tool** The derivative $$f'(x)$$ gives us: * **Slope** of the tangent line at point $$x$$ * **Rate of change** of $$f(x)$$ with respect to $$x$$ * **Instantaneous velocity** if $$f(t)$$ represents position *** ### **2. Tangent and Normal Lines** #### **2.1 Tangent Line** The line that just "touches" a curve at a point, having the same slope as the curve at that point. **Equation at point** $$(x\_0, y\_0)$$**:** $$y - y\_0 = f'(x\_0)(x - x\_0)$$ #### **2.2 Normal Line** The line perpendicular to the tangent line at the point of tangency. **Equation at point** $$(x\_0, y\_0)$$**:** $$y - y\_0 = -\frac{1}{f'(x\_0)}(x - x\_0)$$ (if $$f'(x\_0) \neq 0$$) **Special case:** If $$f'(x\_0) = 0$$, tangent is horizontal, normal is vertical: $$x = x\_0$$ #### **2.3 Example** Find tangent and normal to $$y = x^2$$ at $$(1, 1)$$ $$ f'(x) = 2x \Rightarrow f'(1) = 2 $$ **Tangent line:** $$y - 1 = 2(x - 1) \Rightarrow y = 2x - 1$$ **Normal line:** $$y - 1 = -\frac{1}{2}(x - 1) \Rightarrow y = -\frac{1}{2}x + \frac{3}{2}$$ *** ### **3. Increasing and Decreasing Functions** #### **3.1 Definitions** * **Increasing:** $$f(x\_1) < f(x\_2)$$ whenever $$x\_1 < x\_2$$ * **Decreasing:** $$f(x\_1) > f(x\_2)$$ whenever $$x\_1 < x\_2$$ * **Strictly increasing/decreasing:** No equalities allowed #### **3.2 Test Using Derivatives** * If $$f'(x) > 0$$ on interval $$I$$, then $$f$$ is **increasing** on $$I$$ * If $$f'(x) < 0$$ on interval $$I$$, then $$f$$ is **decreasing** on $$I$$ * If $$f'(x) = 0$$ on interval $$I$$, then $$f$$ is **constant** on $$I$$ #### **3.3 Finding Intervals of Increase/Decrease** **Steps:** 1. Find critical points where $$f'(x) = 0$$ or $$f'(x)$$ undefined 2. Use number line to test sign of $$f'(x)$$ in each interval #### **3.4 Example** Find intervals where $$f(x) = x^3 - 3x^2 - 9x + 5$$ is increasing/decreasing $$ f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1) $$ Critical points: $$x = -1$$, $$x = 3$$ Test intervals: * $$(-\infty, -1)$$**:** Test $$x = -2$$: $$f'(-2) = 3(-5)(-3) = 45 > 0$$ (increasing) * $$(-1, 3)$$**:** Test $$x = 0$$: $$f'(0) = 3(-3)(1) = -9 < 0$$ (decreasing) * $$(3, \infty)$$**:** Test $$x = 4$$: $$f'(4) = 3(1)(5) = 15 > 0$$ (increasing) So: Increasing on $$(-\infty, -1) \cup (3, \infty)$$, decreasing on $$(-1, 3)$$ *** ### **4. Local Maxima and Minima (First Derivative Test)** #### **4.1 Critical Points** A point $$c$$ in domain of $$f$$ is a **critical point** if: 1. $$f'(c) = 0$$, or 2. $$f'(c)$$ does not exist **Important:** All local maxima/minima occur at critical points, but not all critical points give maxima/minima. #### **4.2 First Derivative Test** For critical point $$c$$: 1. If $$f'$$ changes from **positive to negative** at $$c$$: **Local maximum** at $$c$$ 2. If $$f'$$ changes from **negative to positive** at $$c$$: **Local minimum** at $$c$$ 3. If $$f'$$ does **not change sign** at $$c$$: **Neither** maximum nor minimum (inflection point) #### **4.3 Example** Find local extrema of $$f(x) = x^3 - 3x^2 - 9x + 5$$ From previous: $$f'(x) = 3(x-3)(x+1)$$ Critical points: $$x = -1$$, $$x = 3$$ Sign analysis: * Left of $$-1$$: $$f' > 0$$ (increasing) * Right of $$-1$$: $$f' < 0$$ (decreasing) * So $$x = -1$$ is **local maximum** * Left of $$3$$: $$f' < 0$$ (decreasing) * Right of $$3$$: $$f' > 0$$ (increasing) * So $$x = 3$$ is **local minimum** Values: $$f(-1) = 10$$ (local max), $$f(3) = -22$$ (local min) *** ### **5. Concavity and Inflection Points (Second Derivative Test)** #### **5.1 Concavity** * **Concave up:** Graph lies above tangent lines (shaped like ∪) * **Concave down:** Graph lies below tangent lines (shaped like ∩) #### **5.2 Test Using Second Derivative** * If $$f''(x) > 0$$ on interval $$I$$: $$f$$ is **concave up** on $$I$$ * If $$f''(x) < 0$$ on interval $$I$$: $$f$$ is **concave down** on $$I$$ #### **5.3 Inflection Points** A point where concavity changes (from up to down or down to up). **To find inflection points:** 1. Find where $$f''(x) = 0$$ or $$f''(x)$$ undefined 2. Check if concavity changes at those points #### **5.4 Second Derivative Test for Extrema** For critical point $$c$$ with $$f'(c) = 0$$: 1. If $$f''(c) > 0$$: **Local minimum** at $$c$$ 2. If $$f''(c) < 0$$: **Local maximum** at $$c$$ 3. If $$f''(c) = 0$$: **Test inconclusive** (use First Derivative Test) #### **5.5 Example** Analyze $$f(x) = x^4 - 4x^3$$ **First derivative:** $$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$$ Critical points: $$x = 0$$, $$x = 3$$ **Second derivative:** $$f''(x) = 12x^2 - 24x = 12x(x - 2)$$ At $$x = 3$$: $$f''(3) = 12(3)(1) = 36 > 0$$ ⇒ **Local minimum** At $$x = 0$$: $$f''(0) = 0$$ ⇒ Test inconclusive Use First Derivative Test at $$x = 0$$: * Both sides have $$f' < 0$$ (decreasing) * So $$x = 0$$ is **not** extremum **Concavity:** $$f''(x) = 0$$ at $$x = 0$$, $$x = 2$$ Test intervals: * $$(-\infty, 0)$$: Test $$x = -1$$, $$f''(-1) = 12(-1)(-3) = 36 > 0$$ (concave up) * $$(0, 2)$$: Test $$x = 1$$, $$f''(1) = 12(1)(-1) = -12 < 0$$ (concave down) * $$(2, \infty)$$: Test $$x = 3$$, $$f''(3) = 12(3)(1) = 36 > 0$$ (concave up) Inflection points: $$x = 0$$ and $$x = 2$$ *** ### **6. Curve Sketching** #### **6.1 Steps for Curve Sketching** 1. **Domain:** Find where function is defined 2. **Intercepts:** Find x-intercepts ($$f(x)=0$$) and y-intercept ($$f(0)$$) 3. **Symmetry:** Check if even ($$f(-x)=f(x)$$), odd ($$f(-x)=-f(x)$$), or periodic 4. **Asymptotes:** Vertical, horizontal, slant 5. **Intervals of increase/decrease:** Use first derivative 6. **Local extrema:** Identify maxima and minima 7. **Concavity and inflection points:** Use second derivative 8. **Sketch:** Plot key points and connect with appropriate shape #### **6.2 Example: Sketch** $$f(x) = \frac{x^2}{x^2-1}$$ 1. **Domain:** All real $$x$$ except $$x = \pm 1$$ 2. **Intercepts:** $$f(0)=0$$ (y-intercept), $$x=0$$ (x-intercept) 3. **Symmetry:** $$f(-x) = f(x)$$ ⇒ even function (symmetric about y-axis) 4. **Asymptotes:** * Vertical: $$x = 1$$ and $$x = -1$$ (denominator = 0) * Horizontal: As $$x \to \pm\infty$$, $$f(x) \to 1$$ ⇒ $$y=1$$ 5. **First derivative:** $$f'(x) = \frac{2x(x^2-1) - x^2(2x)}{(x^2-1)^2} = \frac{-2x}{(x^2-1)^2}$$ Critical point: $$x=0$$ (where $$f'(x)=0$$) Sign: $$f'(x) > 0$$ for $$x<0$$, $$f'(x) < 0$$ for $$x>0$$ 6. **Increasing:** $$(-\infty, -1) \cup (-1, 0)$$ **Decreasing:** $$(0, 1) \cup (1, \infty)$$ 7. **Local extremum:** Local maximum at $$x=0$$ ($$f(0)=0$$) 8. **Second derivative:** $$f''(x) = \frac{2(3x^2+1)}{(x^2-1)^3}$$ Always positive except at discontinuities ⇒ Always concave up 9. **Sketch:** Symmetric bell shape with vertical asymptotes at ±1, horizontal asymptote at y=1 *** ### **7. Optimization Problems** #### **7.1 Strategy for Optimization** 1. **Understand:** Read problem carefully, identify quantity to optimize 2. **Draw diagram:** If applicable, sketch the situation 3. **Formulate:** Write equation for quantity to optimize (objective function) 4. **Constraint:** Write constraint equation(s) 5. **Reduce variables:** Use constraint to express objective in one variable 6. **Domain:** Determine practical domain 7. **Find critical points:** Take derivative, set to zero 8. **Check endpoints:** Evaluate at critical points and endpoints 9. **Interpret:** Answer in context of problem #### **7.2 Example 1: Maximum Area Problem** Find rectangle of maximum area that can be inscribed in semicircle of radius $$r$$ **Solution:** Let rectangle have width $$2x$$ and height $$y$$ Constraint: $$x^2 + y^2 = r^2$$ (from semicircle equation) Objective: Maximize area $$A = 2xy$$ Express in one variable: $$y = \sqrt{r^2 - x^2}$$ $$A(x) = 2x\sqrt{r^2 - x^2}$$, domain: $$0 \leq x \leq r$$ Find critical points: $$ A'(x) = 2\sqrt{r^2 - x^2} + 2x \cdot \frac{-x}{\sqrt{r^2 - x^2}} = \frac{2(r^2 - x^2) - 2x^2}{\sqrt{r^2 - x^2}} $$ $$ \= \frac{2r^2 - 4x^2}{\sqrt{r^2 - x^2}} $$ Set $$A'(x) = 0$$: $$2r^2 - 4x^2 = 0 \Rightarrow x^2 = \frac{r^2}{2} \Rightarrow x = \frac{r}{\sqrt{2}}$$ Check endpoints: $$A(0) = 0$$, $$A(r) = 0$$ Maximum at $$x = \frac{r}{\sqrt{2}}$$: $$A\_{\text{max}} = 2\left(\frac{r}{\sqrt{2}}\right)\sqrt{r^2 - \frac{r^2}{2}} = r^2$$ #### **7.3 Example 2: Minimize Surface Area** Find dimensions of cylindrical can (closed top and bottom) with volume $$V$$ that minimizes surface area **Solution:** Let radius = $$r$$, height = $$h$$ Volume constraint: $$V = \pi r^2 h \Rightarrow h = \frac{V}{\pi r^2}$$ Surface area: $$S = 2\pi r^2 + 2\pi rh$$ Substitute $$h$$: $$S(r) = 2\pi r^2 + 2\pi r\left(\frac{V}{\pi r^2}\right) = 2\pi r^2 + \frac{2V}{r}$$ Domain: $$r > 0$$ Find critical points: $$ S'(r) = 4\pi r - \frac{2V}{r^2} = 0 $$ $$ 4\pi r = \frac{2V}{r^2} \Rightarrow 4\pi r^3 = 2V \Rightarrow r^3 = \frac{V}{2\pi} \Rightarrow r = \sqrt\[3]{\frac{V}{2\pi}} $$ Then $$h = \frac{V}{\pi r^2} = \frac{V}{\pi\left(\frac{V}{2\pi}\right)^{2/3}} = \frac{V^{1/3}}{(\pi)^{1/3}}\left(\frac{2\pi}{V}\right)^{2/3} = 2\sqrt\[3]{\frac{V}{2\pi}} = 2r$$ Optimal ratio: $$h = 2r$$ (height = diameter) *** ### **8. Related Rates** #### **8.1 Concept** Problems where two or more related quantities are changing with time, and we want to find how fast one is changing given how fast others are changing. #### **8.2 Strategy** 1. **Identify variables:** What quantities are changing? Assign variables 2. **Given rates:** What rates are known? $$\frac{d?}{dt} = ?$$ 3. **Wanted rate:** What rate do we need to find? $$\frac{d?}{dt} = ?$$ 4. **Equation:** Write equation relating variables 5. **Differentiate:** Differentiate with respect to time $$t$$ 6. **Substitute:** Plug in known values 7. **Solve:** Solve for wanted rate #### **8.3 Example 1: Ladder Problem** A 10 ft ladder leans against wall. Bottom slides away at 1 ft/s. How fast is top sliding down when bottom is 6 ft from wall? **Solution:** Let $$x$$ = distance from wall, $$y$$ = height on wall Given: $$\frac{dx}{dt} = 1$$ ft/s Find: $$\frac{dy}{dt}$$ when $$x = 6$$ Equation: $$x^2 + y^2 = 100$$ Differentiate: $$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$ When $$x = 6$$: $$y = \sqrt{100 - 36} = 8$$ Substitute: $$2(6)(1) + 2(8)\frac{dy}{dt} = 0$$ $$12 + 16\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{3}{4}$$ ft/s Negative means sliding down. #### **8.4 Example 2: Spherical Balloon** Air is pumped into spherical balloon at rate of 100 cm³/s. How fast is radius increasing when radius is 10 cm? **Solution:** Given: $$\frac{dV}{dt} = 100$$ cm³/s Find: $$\frac{dr}{dt}$$ when $$r = 10$$ Volume: $$V = \frac{4}{3}\pi r^3$$ Differentiate: $$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$ When $$r = 10$$: $$100 = 4\pi(10)^2 \frac{dr}{dt} = 400\pi \frac{dr}{dt}$$ $$\frac{dr}{dt} = \frac{100}{400\pi} = \frac{1}{4\pi} \approx 0.0796$$ cm/s *** ### **9. Mean Value Theorems** #### **9.1 Rolle's Theorem** If $$f$$ is: 1. Continuous on $$\[a,b]$$ 2. Differentiable on $$(a,b)$$ 3. $$f(a) = f(b)$$ Then there exists $$c$$ in $$(a,b)$$ such that $$f'(c) = 0$$ **Geometric meaning:** Between two points with same height, there's a horizontal tangent. #### **9.2 Mean Value Theorem (MVT)** If $$f$$ is: 1. Continuous on $$\[a,b]$$ 2. Differentiable on $$(a,b)$$ Then there exists $$c$$ in $$(a,b)$$ such that: $$ f'(c) = \frac{f(b) - f(a)}{b - a} $$ **Geometric meaning:** There's a point where tangent slope equals secant slope. #### **9.3 Example Using MVT** Verify MVT for $$f(x) = x^3 - x$$ on $$\[0, 2]$$ **Check conditions:** Polynomial ⇒ continuous and differentiable everywhere $$ \frac{f(2) - f(0)}{2 - 0} = \frac{(8-2) - 0}{2} = \frac{6}{2} = 3 $$ Find $$c$$ such that $$f'(c) = 3$$ $$f'(x) = 3x^2 - 1$$ Set $$3x^2 - 1 = 3 \Rightarrow 3x^2 = 4 \Rightarrow x^2 = \frac{4}{3} \Rightarrow x = \pm\frac{2}{\sqrt{3}}$$ In $$(0,2)$$: $$c = \frac{2}{\sqrt{3}} \approx 1.155$$ #### **9.4 Consequences of MVT** 1. If $$f'(x) = 0$$ for all $$x$$ in $$(a,b)$$, then $$f$$ is constant on $$(a,b)$$ 2. If $$f'(x) > 0$$ for all $$x$$ in $$(a,b)$$, then $$f$$ is increasing on $$(a,b)$$ 3. If $$f'(x) < 0$$ for all $$x$$ in $$(a,b)$$, then $$f$$ is decreasing on $$(a,b)$$ *** ### **10. Linear Approximation and Differentials** #### **10.1 Linear Approximation** Approximate function near point $$a$$ using tangent line: $$ L(x) = f(a) + f'(a)(x - a) $$ Approximation: $$f(x) \approx L(x)$$ for $$x$$ near $$a$$ #### **10.2 Example** Approximate $$\sqrt{16.1}$$ using linear approximation Let $$f(x) = \sqrt{x}$$, $$a = 16$$ $$f(a) = \sqrt{16} = 4$$ $$f'(x) = \frac{1}{2\sqrt{x}} \Rightarrow f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{8} = 0.125$$ Linear approximation: $$L(x) = 4 + 0.125(x - 16)$$ For $$x = 16.1$$: $$\sqrt{16.1} \approx 4 + 0.125(0.1) = 4 + 0.0125 = 4.0125$$ Actual: $$\sqrt{16.1} \approx 4.01248$$ (very close!) #### **10.3 Differentials** If $$y = f(x)$$, then differential $$dy = f'(x) dx$$ **Interpretation:** $$dy$$ approximates change in $$y$$ when $$x$$ changes by $$dx$$ **Example:** Use differentials to estimate change in volume of sphere when radius increases from 10 cm to 10.1 cm Volume: $$V = \frac{4}{3}\pi r^3$$ $$dV = 4\pi r^2 dr$$ Here $$r = 10$$, $$dr = 0.1$$ $$dV = 4\pi(10)^2(0.1) = 40\pi \approx 125.66$$ cm³ Actual change: $$\Delta V = \frac{4}{3}\pi(10.1^3 - 10^3) = \frac{4}{3}\pi(1030.301 - 1000) \approx 127.17$$ cm³ *** ### **11. L'Hôpital's Rule** #### **11.1 Statement** If $$\lim\_{x \to a} \frac{f(x)}{g(x)}$$ has form $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}$$, then: $$ \lim\_{x \to a} \frac{f(x)}{g(x)} = \lim\_{x \to a} \frac{f'(x)}{g'(x)} $$ provided the limit on right exists or is $$\pm\infty$$ #### **11.2 Examples** **Example 1:** $$\lim\_{x \to 0} \frac{\sin x}{x}$$ ($$\frac{0}{0}$$ form) Apply L'Hôpital's: $$\lim\_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1$$ **Example 2:** $$\lim\_{x \to \infty} \frac{x^2}{e^x}$$ ($$\frac{\infty}{\infty}$$ form) Apply L'Hôpital's twice: First: $$\lim\_{x \to \infty} \frac{2x}{e^x}$$ Second: $$\lim\_{x \to \infty} \frac{2}{e^x} = 0$$ **Example 3:** $$\lim\_{x \to 0^+} x \ln x$$ ($$0 \cdot \infty$$ form) Rewrite: $$\lim\_{x \to 0^+} \frac{\ln x}{1/x}$$ (now $$\frac{\infty}{\infty}$$) Apply L'Hôpital's: $$\lim\_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim\_{x \to 0^+} (-x) = 0$$ #### **11.3 Other Indeterminate Forms** Can convert to $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}$$: 1. $$0 \cdot \infty$$: Write as $$\frac{0}{1/\infty}$$ or $$\frac{\infty}{1/0}$$ 2. $$\infty - \infty$$: Combine into single fraction 3. $$0^0$$, $$\infty^0$$, $$1^\infty$$: Take natural log *** ### **12. Newton's Method for Finding Roots** #### **12.1 The Method** Iterative method to approximate roots of $$f(x) = 0$$ **Formula:** $$ x\_{n+1} = x\_n - \frac{f(x\_n)}{f'(x\_n)} $$ #### **12.2 Steps** 1. Choose initial approximation $$x\_0$$ 2. Compute $$x\_1 = x\_0 - \frac{f(x\_0)}{f'(x\_0)}$$ 3. Repeat until desired accuracy #### **12.3 Example** Find approximate root of $$f(x) = x^2 - 2 = 0$$ (i.e., $$\sqrt{2}$$) $$f'(x) = 2x$$ Choose $$x\_0 = 1$$ Iteration 1: $$x\_1 = 1 - \frac{1^2 - 2}{2(1)} = 1 - \frac{-1}{2} = 1.5$$ Iteration 2: $$x\_2 = 1.5 - \frac{(1.5)^2 - 2}{2(1.5)} = 1.5 - \frac{2.25 - 2}{3} = 1.5 - \frac{0.25}{3} \approx 1.4167$$ Iteration 3: $$x\_3 = 1.4167 - \frac{(1.4167)^2 - 2}{2(1.4167)} \approx 1.4142$$ $$\sqrt{2} \approx 1.41421356$$, so we're already close! *** ### **13. Applications in Physics and Economics** #### **13.1 Physics Applications** **Motion along a line:** * Position: $$s(t)$$ * Velocity: $$v(t) = s'(t)$$ * Acceleration: $$a(t) = v'(t) = s''(t)$$ **Example:** Particle moves with $$s(t) = t^3 - 6t^2 + 9t$$ * Velocity: $$v(t) = 3t^2 - 12t + 9$$ * Acceleration: $$a(t) = 6t - 12$$ When is particle at rest? $$v(t) = 0 \Rightarrow 3(t^2 - 4t + 3) = 0 \Rightarrow t = 1, 3$$ **Related rates in physics:** * Pressure and volume: Boyle's Law $$PV = k$$ * Force and work: $$W = F \cdot d$$ #### **13.2 Economics Applications** **Marginal Concepts:** * **Marginal cost:** $$C'(x)$$ ≈ cost of producing one more unit * **Marginal revenue:** $$R'(x)$$ ≈ revenue from selling one more unit * **Marginal profit:** $$P'(x) = R'(x) - C'(x)$$ **Elasticity of Demand:** $$ E(p) = \frac{p}{q} \cdot \frac{dq}{dp} $$ Interpretation: * $$|E| > 1$$: Elastic (demand sensitive to price) * $$|E| = 1$$: Unit elastic * $$|E| < 1$$: Inelastic (demand not sensitive to price) **Example:** Demand $$q = 100 - 2p$$ $$ \frac{dq}{dp} = -2 $$ $$ E(p) = \frac{p}{100 - 2p} \cdot (-2) = -\frac{2p}{100 - 2p} $$ At $$p = 20$$: $$E(20) = -\frac{40}{60} = -\frac{2}{3} \Rightarrow |E| = \frac{2}{3} < 1$$ (inelastic) *** ### **14. Important Theorems and Formulas** #### **14.1 Key Theorems** 1. **Rolle's Theorem:** $$f(a)=f(b) \Rightarrow \exists c: f'(c)=0$$ 2. **Mean Value Theorem:** $$\exists c: f'(c) = \frac{f(b)-f(a)}{b-a}$$ 3. **First Derivative Test:** Sign change of $$f'$$ indicates local extrema 4. **Second Derivative Test:** $$f''(c)$$ sign indicates concavity and extremum type #### **14.2 Optimization Formulas** 1. **Rectangle area:** $$A = lw$$, perimeter $$P = 2l + 2w$$ 2. **Cylinder volume:** $$V = \pi r^2 h$$, surface area $$S = 2\pi r^2 + 2\pi rh$$ 3. **Sphere volume:** $$V = \frac{4}{3}\pi r^3$$, surface area $$S = 4\pi r^2$$ 4. **Cone volume:** $$V = \frac{1}{3}\pi r^2 h$$ #### **14.3 Related Rates Common Formulas** 1. Pythagorean: $$x^2 + y^2 = z^2$$ 2. Similar triangles: $$\frac{a}{b} = \frac{c}{d}$$ 3. Volume formulas (see above) 4. Trigonometric relations in right triangles *** ### **15. Solved Examples** #### **Example 1: Complete Curve Analysis** Analyze $$f(x) = \frac{x}{x^2+1}$$ 1. **Domain:** All real numbers (denominator never 0) 2. **Intercepts:** $$f(0)=0$$, only intercept 3. **Symmetry:** $$f(-x) = -f(x)$$ ⇒ odd function (symmetric about origin) 4. **Asymptotes:** * Horizontal: As $$x \to \pm\infty$$, $$f(x) \to 0$$ ⇒ $$y=0$$ * No vertical asymptotes 5. **First derivative:** $$f'(x) = \frac{(1)(x^2+1) - x(2x)}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}$$ Critical points: $$f'(x)=0 \Rightarrow 1-x^2=0 \Rightarrow x=\pm 1$$ Sign: * $$x<-1$$: $$f' < 0$$ (decreasing) * $$-1\ 0$$ (increasing) * $$x>1$$: $$f' < 0$$ (decreasing) 6. **Extrema:** Local min at $$x=-1$$ ($$f(-1)=-\frac{1}{2}$$), local max at $$x=1$$ ($$f(1)=\frac{1}{2}$$) 7. **Second derivative:** $$f''(x) = \frac{2x(x^2-3)}{(x^2+1)^3}$$ Inflection points: $$x=0$$, $$x=\pm\sqrt{3}$$ 8. **Sketch:** S-shaped curve through origin, max at (1, 1/2), min at (-1, -1/2) #### **Example 2: Optimization** Find point on parabola $$y^2 = 2x$$ closest to point (1, 4) **Solution:** Distance from (x,y) to (1,4): $$D = \sqrt{(x-1)^2 + (y-4)^2}$$ Constraint: $$y^2 = 2x \Rightarrow x = \frac{y^2}{2}$$ Minimize $$D^2$$ (easier): $$f(y) = \left(\frac{y^2}{2}-1\right)^2 + (y-4)^2$$ $$f'(y) = 2\left(\frac{y^2}{2}-1\right)y + 2(y-4) = y^3 - 2y + 2y - 8 = y^3 - 8$$ Set $$f'(y)=0$$: $$y^3=8 \Rightarrow y=2$$ Then $$x = \frac{y^2}{2} = \frac{4}{2} = 2$$ Closest point: (2, 2) Distance: $$D = \sqrt{(2-1)^2 + (2-4)^2} = \sqrt{1+4} = \sqrt{5}$$ *** ### **16. Common Mistakes and Exam Tips** #### **16.1 Common Mistakes** 1. **Forgetting to check endpoints** in optimization problems 2. **Misapplying chain rule** in related rates 3. **Confusing average rate of change** with instantaneous rate 4. **Not verifying conditions** for Rolle's/MVT 5. **Sign errors** in derivative calculations #### **16.2 Problem-Solving Strategy** 1. **Read carefully:** Understand what's given and what's asked 2. **Draw diagram:** Especially for geometry/optimization problems 3. **Define variables:** Clearly label all quantities 4. **Write equations:** Relate variables mathematically 5. **Differentiate correctly:** Watch for chain rule, product rule, etc. 6. **Check units:** Ensure consistency in related rates 7. **Verify answer:** Does it make sense in context? #### **16.3 Quick Checks** 1. **Local max/min:** $$f'$$ changes sign at critical point 2. **Inflection point:** $$f''$$ changes sign 3. **Tangent line:** $$y - y\_0 = f'(x\_0)(x - x\_0)$$ 4. **Related rates:** All derivatives are with respect to same variable (usually time) 5. **Optimization:** Check both critical points and endpoints This comprehensive theory covers all applications of derivatives with detailed explanations and examples, providing complete preparation for the entrance examination.