Introduction to Differentiation

Fundamental Concepts of Differentiation What Is a Derivative? The derivative measures how quickly a function is changing at a specific point. Rather than asking "how much did the function change over an interval?" (which gives an average rate of change), the derivative asks "how fast is the function changing right now at this exact point?" (instantaneous rate of change). Mathematically, the derivative of a function $f$ at a point $a$ is defined as: $$f'(a) = \lim{h \to 0} \frac{f(a+h) - f(a)}{h}$$ This expression is called the difference quotient. The numerator measures how much the function value changed, and the denominator measures how much the input changed. The key is that we take the limit as $h \to 0$—we shrink the input interval to an infinitesimal size to capture the instantaneous rate of change. Important: The limit must exist for the derivative to be defined at that point. If the limit doesn't exist, the function is not differentiable at that point. Geometric Meaning Geometrically, the derivative $f'(a)$ gives the slope of the tangent line to the graph of $f$ at the point $(a, f(a))$. The difference quotient represents the slope of a secant line through $(a, f(a))$ and $(a+h, f(a+h))$. As we shrink $h$ to zero, the secant line approaches the tangent line, and the difference quotient approaches the slope of that tangent. From Average Change to Instantaneous Change Understanding the distinction between average and instantaneous rates of change is crucial. Average rate of change over an interval $[a, a+h]$ is straightforward: it's just $\frac{f(a+h) - f(a)}{h}$. This tells you how much the function changed per unit of input over that interval. If you drive 120 miles in 2 hours, your average rate is 60 miles per hour—but you may have gone faster or slower at different moments. Instantaneous rate of change captures the rate at a single moment. To find it, we take the average rate over a shrinking interval: as $h \to 0$, the interval becomes so small that it essentially tells us the rate at the single point $a$. This is what the limit achieves. The transition from a finite interval (average) to an infinitesimal one (instantaneous) is what makes limits the gateway to calculus. Without limits, we can only describe how functions behave over intervals; with limits, we can describe how they behave at individual points. The Role of Limits in Differentiation Limits solve a subtle problem: the difference quotient $\frac{f(a+h) - f(a)}{h}$ is undefined when $h = 0$ (you get $\frac{0}{0}$). We can't just plug in $h = 0$ directly. However, we can evaluate what happens as $h$ gets arbitrarily close to zero. The limit $\lim{h \to 0}$ captures this idea: we examine the behavior of the difference quotient for smaller and smaller values of $h$, and determine what value the quotient approaches. If it approaches a specific number, that number is the derivative. This approach eliminates the dependence on any particular interval length. Instead of saying "the rate of change over an interval of length 0.1 is approximately...", the limit gives us the exact instantaneous rate at the point. When Does a Derivative Not Exist? A function may fail to be differentiable at certain points, even if the function is continuous there. Watch out for these cases: Corners or sharp points: If the graph has a sharp corner, the slope changes abruptly and no single tangent line exists. Example: $f(x) = |x|$ is continuous at $x = 0$ but not differentiable there. Cusps: Similar to corners, but where the graph comes to a sharp point. The left and right derivatives don't match. Vertical tangent lines: If the graph has a vertical tangent at a point, the slope is undefined (it's $\infty$). Discontinuities: If the function jumps, has a hole, or has any break at a point, it cannot be differentiable there. Key insight: Continuity is necessary for differentiability, but not sufficient. A function must be continuous at a point to be differentiable there, but continuity alone doesn't guarantee differentiability—the absolute value function shows this. Basic Differentiation Rules Computing derivatives directly from the limit definition is tedious. Instead, we use derivative rules that allow us to quickly find derivatives of combinations of simpler functions. The Power Rule The most fundamental rule is the power rule: $$\frac{d}{dx}(x^n) = n \, x^{n-1}$$ where $n$ is any real number (integer, fraction, negative, etc.). Examples: $\frac{d}{dx}(x^3) = 3x^2$ $\frac{d}{dx}(x^{-2}) = -2x^{-3}$ $\frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$ The power rule works because when you expand the difference quotient for $x^n$ and simplify, the limit gives you $n \, x^{n-1}$. The Constant-Multiple Rule If you multiply a function by a constant $c$, you also multiply its derivative by that constant: $$\frac{d}{dx}(c \cdot f(x)) = c \cdot f'(x)$$ Example: $\frac{d}{dx}(5x^2) = 5 \cdot \frac{d}{dx}(x^2) = 5 \cdot 2x = 10x$ This rule is intuitive: multiplying the function vertically by a constant stretches the slope by that same factor. The Sum Rule The derivative of a sum equals the sum of the derivatives: $$\frac{d}{dx}(f(x) + g(x)) = f'(x) + g'(x)$$ Example: $\frac{d}{dx}(x^3 + x^2) = 3x^2 + 2x$ This rule extends to sums of any number of terms. Combined with the power rule and constant-multiple rule, you can differentiate any polynomial. The Product Rule When multiplying two functions, you cannot simply multiply their derivatives. Instead, use the product rule: $$\frac{d}{dx}(f(x) \cdot g(x)) = f'(x) \cdot g(x) + f(x) \cdot g'(x)$$ A helpful mnemonic: "first times derivative of second, plus second times derivative of first." Example: Find $\frac{d}{dx}((x^2)(x^3 + 1))$. Let $f(x) = x^2$ and $g(x) = x^3 + 1$. Then $f'(x) = 2x$ and $g'(x) = 3x^2$. $$\frac{d}{dx}((x^2)(x^3 + 1)) = (2x)(x^3 + 1) + (x^2)(3x^2) = 2x^4 + 2x + 3x^4 = 5x^4 + 2x$$ Why not just multiply the derivatives? If you tried $f'(x) \cdot g'(x) = 2x \cdot 3x^2 = 6x^3$, you'd get the wrong answer. The product rule accounts for the fact that both functions are changing simultaneously. The Quotient Rule When dividing two functions, use the quotient rule: $$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}$$ provided $g(x) \neq 0$. The mnemonic is: "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared." Example: Find $\frac{d}{dx}\left(\frac{x^2}{x + 1}\right)$. Let $f(x) = x^2$ and $g(x) = x + 1$. Then $f'(x) = 2x$ and $g'(x) = 1$. $$\frac{d}{dx}\left(\frac{x^2}{x + 1}\right) = \frac{(2x)(x + 1) - (x^2)(1)}{(x+1)^2} = \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}$$ Watch out: The order matters in the numerator. It's "bottom times derivative of top" minus "top times derivative of bottom," not the other way around. Reversing the order will give you the opposite sign. Advanced Differentiation Rules The Chain Rule for Composite Functions Many functions are composed of other functions. For example, $\sqrt{x^2 + 1}$ is the square root function applied to $x^2 + 1$. The chain rule tells us how to differentiate these composites. If $h(x) = g(f(x))$ (meaning $f$ is applied first, then $g$), then: $$h'(x) = g'(f(x)) \cdot f'(x)$$ In other words: take the derivative of the outer function, evaluate it at the inner function, then multiply by the derivative of the inner function. Example: Find $\frac{d}{dx}(\sin(x^2))$. The outer function is $\sin$ and the inner function is $x^2$. Derivative of the outer function (sine) is $\cos$. Evaluated at the inner function: $\cos(x^2)$. Derivative of the inner function: $2x$. Result: $\frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot 2x = 2x\cos(x^2)$. Why do we need it? Without the chain rule, you might incorrectly compute the derivative of $\sin(x^2)$ as $\cos(x^2)$ (forgetting to account for the change in $x^2$). The chain rule ensures we account for how the inner function is changing. The chain rule is one of the most important and most frequently used rules in calculus. Mastering it is essential. Implicit Differentiation Sometimes a relationship between $x$ and $y$ is given implicitly—as an equation like $x^2 + y^2 = 25$ rather than explicitly as $y = f(x)$. Even so, we can find $\frac{dy}{dx}$ using implicit differentiation. The strategy is: Differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$. When differentiating terms involving $y$, use the chain rule (remember that $\frac{dy}{dx}$ appears whenever you differentiate $y$). Solve the resulting equation for $\frac{dy}{dx}$. Example: Find $\frac{dy}{dx}$ for $x^2 + y^2 = 25$. Differentiate both sides with respect to $x$: $$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$$ $$2x + 2y \frac{dy}{dx} = 0$$ Solve for $\frac{dy}{dx}$: $$2y \frac{dy}{dx} = -2x$$ $$\frac{dy}{dx} = -\frac{x}{y}$$ Notice that the derivative is expressed in terms of both $x$ and $y$, rather than just $x$ alone. This is typical for implicit differentiation. <extrainfo> Logarithmic Differentiation Logarithmic differentiation is a technique that simplifies the derivative of complicated products, quotients, or powers. The idea is to take the natural logarithm of both sides of an equation, differentiate, and then solve. Strategy: Start with $y = f(x)$ where $f(x)$ is complicated. Take the natural logarithm of both sides: $\ln(y) = \ln(f(x))$. Use logarithm properties to simplify the right side. Differentiate both sides implicitly with respect to $x$. Solve for $\frac{dy}{dx}$. Example: Find $\frac{d}{dx}(x^x)$ using logarithmic differentiation. Let $y = x^x$. Then $\ln(y) = \ln(x^x) = x \ln(x)$. Differentiate both sides: $$\frac{1}{y} \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$$ Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = y(\ln(x) + 1) = x^x(\ln(x) + 1)$$ This technique is particularly useful when the variable appears in both the base and exponent. </extrainfo> <extrainfo> Higher-Order Derivatives The derivative of a function is itself a function, so we can take its derivative as well. The second derivative, denoted $f''(x)$ or $\frac{d^2f}{dx^2}$, represents the rate of change of the first derivative. Similarly, the third derivative is $f'''(x)$, and so on. Example: If $f(x) = x^3 + 2x^2$, then: First derivative: $f'(x) = 3x^2 + 4x$ Second derivative: $f''(x) = 6x + 4$ Third derivative: $f'''(x) = 6$ Fourth derivative: $f^{(4)}(x) = 0$ Higher-order derivatives have interpretations: in physics, if position is $f(t)$, then $f'(t)$ is velocity and $f''(t)$ is acceleration. </extrainfo> Standard Derivative Formulas Beyond algebraic functions, you need to memorize the derivatives of common transcendental functions (trigonometric, exponential, and logarithmic). Trigonometric Functions The derivatives of the basic trigonometric functions are: $$\frac{d}{dx}(\sin(x)) = \cos(x)$$ $$\frac{d}{dx}(\cos(x)) = -\sin(x)$$ $$\frac{d}{dx}(\tan(x)) = \sec^2(x)$$ Memory tricks: Sine and cosine derivatives are related and opposite in sign (cosine is negative). The tangent derivative involves secant (the reciprocal of cosine), squared. Example: $\frac{d}{dx}(\sin(3x))$ requires the chain rule: $\cos(3x) \cdot 3 = 3\cos(3x)$. Exponential Functions The exponential function $e^x$ has a remarkable property: its derivative is itself. $$\frac{d}{dx}(e^x) = e^x$$ For exponential functions with other bases: $$\frac{d}{dx}(a^x) = a^x \ln(a)$$ where $a > 0$ and $a \neq 1$. Example: $\frac{d}{dx}(e^{2x}) = e^{2x} \cdot 2 = 2e^{2x}$ (using the chain rule) $\frac{d}{dx}(2^x) = 2^x \ln(2)$ The presence of $\ln(a)$ appears because the base is not $e$; if you use base $e$, that factor becomes $\ln(e) = 1$. Logarithmic Functions The natural logarithm and logarithms with other bases have these derivatives: $$\frac{d}{dx}(\ln(x)) = \frac{1}{x}$$ $$\frac{d}{dx}(\loga(x)) = \frac{1}{x \ln(a)}$$ where $a > 0$ and $a \neq 1$. Example: $\frac{d}{dx}(\ln(x^2)) = \frac{1}{x^2} \cdot 2x = \frac{2}{x}$ (using the chain rule) $\frac{d}{dx}(\log{10}(x)) = \frac{1}{x \ln(10)}$ Notice that the derivative of $\ln(x)$ is remarkably simple. This is one reason the natural logarithm is preferred in calculus. Interpretations and Applications of the Derivative Physical Interpretation: Position, Velocity, and Acceleration Derivatives appear naturally in physics. If $s(t)$ represents the position of an object at time $t$, then: Velocity is $v(t) = s'(t)$, the instantaneous rate of change of position. Acceleration is $a(t) = v'(t) = s''(t)$, the instantaneous rate of change of velocity. Example: If a ball is thrown upward and its height after $t$ seconds is $s(t) = 64t - 16t^2$ feet, then: Velocity: $v(t) = 64 - 32t$ feet per second Acceleration: $a(t) = -32$ feet per second squared (constant, due to gravity) At $t = 1$ second, the velocity is $v(1) = 64 - 32 = 32$ feet per second. At $t = 2$ seconds, $v(2) = 0$ feet per second—the ball momentarily stops at its peak before falling back down. Role in Optimization: Finding Maxima and Minima The derivative reveals where a function increases and decreases, which is essential for optimization. If $f'(x) > 0$ on an interval, the function is increasing there. If $f'(x) < 0$ on an interval, the function is decreasing there. If $f'(x) = 0$ at a point, the function has a horizontal tangent—a critical point. Critical points are candidates for local maxima and minima. To find the maximum or minimum value of a function on an interval, evaluate the function at: All critical points (where $f'(x) = 0$ or $f'$ is undefined) The endpoints of the interval Then compare these values. Example: Find the maximum of $f(x) = 12x - x^2$ on $[0, 12]$. First, find critical points: $f'(x) = 12 - 2x = 0 \Rightarrow x = 6$. Evaluate at the critical point and endpoints: $f(0) = 0$ $f(6) = 12(6) - 6^2 = 72 - 36 = 36$ $f(12) = 12(12) - 12^2 = 144 - 144 = 0$ The maximum value is $36$, achieved at $x = 6$. Connection to Integral Calculus The Fundamental Theorem of Calculus establishes that differentiation and integration are inverse operations. If you differentiate a function and then integrate the result, you recover the original function (up to a constant). Conversely, if you integrate a function and then differentiate, you get back the original. This deep connection means that understanding derivatives gives you insight into integrals, and vice versa. While this course focuses on differentiation, knowing that integration "undoes" differentiation is valuable context.