No Surprises Allowed Continuity and Differentiability

Take a pen, put its tip anywhere on the graph of a function, and trace the curve from left to right without lifting the pen. If you can trace the whole graph that way, without ever having to pick the pen up and set it down somewhere else, the function is continuous. If you are forced to lift the pen even once, the function fails to be continuous at that spot. This is a genuinely useful intuition, but it is not a definition you can use to prove anything, so mathematics tightens it into three checkable conditions. A function f is continuous at a point x = a exactly when all three of the following hold: first, f(a) itself is defined (there is actually a point to draw); second, lim(x to a) f(x) exists (the curve is heading toward some single value from both sides); and third, that limit actually equals f(a) (the curve arrives exactly where the point sits, with no gap between them). Fail any one of the three, and continuity breaks. Consider f(x) = (x squared - 1)/(x - 1) for x not equal to 1, with f(1) separately defined as 5. As x approaches 1, f(x) behaves like (x+1), since the (x-1) factors cancel, so lim(x to 1) f(x) = 2. But f(1) has been separately set to 5. The limit exists, and f(1) is defined, but they disagree, 2 against 5, so f fails the third condition and is not continuous at x = 1, even though nothing about the function looks dramatic there. This is exactly the pen-lifting picture: you would trace the curve smoothly toward the point (1, 2), then have to jump the pen up to the isolated point (1, 5) sitting above it.

Checking all three conditions from scratch for every function you meet would be exhausting. Fortunately, continuity behaves well under the ordinary operations of algebra. If f and g are both continuous at a point, then so are f+g, f-g, and f times g, and so is f/g wherever g is not zero there. This means you never actually need to check the three conditions for something like h(x) = x squared + sin(x), because x squared and sin(x) are each already known to be continuous everywhere (a polynomial has no gaps or jumps anywhere, and neither does a smooth wave like sine), and the sum of two continuous functions is automatically continuous. The same protection extends to composition: if g is continuous at a, and f is continuous at g(a), then the composite function f(g(x)) is continuous at a too. Composing continuous functions can never manufacture a break out of nowhere. Between the algebra rules and the composition rule, an enormous share of the functions that show up in this course, polynomials, trigonometric functions, and combinations of them built by adding, multiplying, dividing, or nesting, are continuous everywhere they are defined, and 'everywhere they are defined' usually just means avoiding a handful of specific points where a denominator would hit zero.

Differentiability is a stricter demand than continuity. A function is differentiable at x = a when f prime(a) = lim(h to 0) [f(a+h) - f(a)]/h actually exists as a single number, meaning the curve has one well defined tangent line slope right at that point, not two different slopes depending on which side you approach from. Every differentiable function is automatically continuous, and this direction can be proved directly: if the limit defining f prime(a) exists, a short algebraic rearrangement shows lim(h to 0) [f(a+h) - f(a)] must equal 0, which is exactly what continuity at a requires. But the reverse implication fails, and the standard example is f(x) = |x| at x = 0. This function is perfectly continuous there, no break, no jump, f(0) = 0 and the graph passes right through the origin without lifting the pen. Yet it is not differentiable at x = 0. Approach from the right (h greater than 0): [f(0+h) - f(0)]/h = |h|/h = h/h = 1. Approach from the left (h less than 0): [f(0+h) - f(0)]/h = |h|/h = (-h)/h = -1. The right hand derivative is 1, the left hand derivative is -1, and since they disagree, f prime(0) does not exist at all. Graphically, this is the sharp corner at the bottom of the V-shaped graph of |x|: the curve is unbroken, but it kinks, and a kink has no single well defined tangent line slope. Smoothness is a strictly stronger property than merely being unbroken.

Many functions worth differentiating are not single, simple expressions but one function tucked inside another, like sin(x squared), or (3x + 1) to the power 5. Call the outer function f and the inner function g, so the whole thing is f(g(x)). The chain rule says: differentiate the outer function first, leaving the inner function untouched inside it, then multiply by the derivative of the inner function. In symbols, d/dx [f(g(x))] = f prime(g(x)) times g prime(x). Take y = sin(x squared). The outer function is sine, whose derivative is cosine, so differentiating the outer layer while leaving x squared alone inside gives cos(x squared). Then multiply by the derivative of the inner function, g(x) = x squared, which is 2x. The result: dy/dx = cos(x squared) times 2x = 2x cos(x squared). Try (3x+1) to the power 5. Outer function: something to the power 5, whose derivative (by the power rule) is 5 times something to the power 4, giving 5(3x+1) to the power 4, leaving the inner (3x+1) untouched. Multiply by the derivative of the inner function 3x+1, which is 3. Result: dy/dx = 15(3x+1) to the power 4. The chain rule is used so constantly, in every remaining chapter of this thread, that it is worth practicing until peeling the outer layer off first becomes automatic.

Every derivative so far has come from a function written explicitly as y = (some expression in x). But many important curves are given by an equation mixing x and y together with no way to isolate y at all, like the circle x squared + y squared = 25. Implicit differentiation handles this by differentiating both sides of the equation with respect to x, treating y throughout as a hidden function of x, and applying the chain rule every time y appears. Differentiate x squared + y squared = 25 term by term. The derivative of x squared is 2x, ordinary power rule. The derivative of y squared, since y is secretly a function of x, needs the chain rule: treat y squared as (something) squared, whose derivative is 2 times that something, times the derivative of that something with respect to x, giving 2y times dy/dx. The derivative of the constant 25 is 0. So 2x + 2y(dy/dx) = 0. Solve for dy/dx by isolating it: 2y(dy/dx) = -2x, so dy/dx = -x/y. This single formula gives the slope of the circle at any point on it. At the point (3, 4), the slope is -3/4; at (0, 5), the slope is -0/5 = 0, exactly right for the flat top of the circle. The trick, every single time, is remembering that y is not just another variable like x; it carries its own hidden dy/dx along whenever the chain rule reaches it.

Implicit differentiation also unlocks the derivatives of the inverse trigonometric functions. Let y = sin inverse(x), meaning sin(y) = x, with y restricted to the range where sine is one-to-one. Differentiate both sides with respect to x, treating y as hidden: cos(y) times dy/dx = 1, so dy/dx = 1/cos(y). To turn this into an expression purely in x, use the identity sin squared(y) + cos squared(y) = 1, so cos(y) = the square root of (1 - sin squared(y)) = the square root of (1 - x squared), since sin(y) = x by definition. Substituting back gives d/dx[sin inverse(x)] = 1 divided by the square root of (1 - x squared). The remaining five inverse trig derivatives follow the same method, and are usually simply memorised as a table once derived: the derivative of cos inverse(x) is negative 1 over the square root of (1 - x squared), exactly the negative of sin inverse's derivative, since cos inverse(x) and sin inverse(x) always add up to the constant pi/2. The derivative of tan inverse(x) is 1/(1 + x squared). Notice something striking about all of these: none of them contain a single trigonometric function in the final answer. Differentiating an inverse trig function produces a purely algebraic expression, built only from x, square roots, and fractions.

Two more standard derivatives complete the toolkit: d/dx(e to the x) = e to the x, meaning the natural exponential function is its own derivative, unique among all functions in that respect, and d/dx(log x) = 1/x (using natural logarithm throughout). Both results rely on a careful limit argument involving how e itself is defined, detailed enough that even this textbook sets the full proof aside into supplementary material rather than working through it in the main chapter, so this chapter states both results and puts them straight to use. They enable a technique called logarithmic differentiation, useful whenever a variable sits in an exponent, such as y = x to the power x, a function the power rule and the exponential rule cannot handle alone, since the base and the exponent are both x. Take the natural log of both sides first: log(y) = x times log(x), using the log rule that log(a to the power b) = b times log(a). Now differentiate both sides with respect to x. The left side needs the chain rule: d/dx[log(y)] = (1/y) times dy/dx. The right side needs the product rule, since it is x times log(x): d/dx[x log(x)] = 1 times log(x) + x times (1/x) = log(x) + 1. So (1/y)(dy/dx) = log(x) + 1, giving dy/dx = y times [log(x) + 1] = x to the power x times [log(x) + 1], substituting the original y back in at the end. Logarithmic differentiation turns an otherwise impossible exponent-of-a-variable problem into an ordinary product-rule problem.

Some curves are most naturally described not as y in terms of x, but with both x and y written in terms of a third variable, a parameter, often called t, as in x = a cos(t), y = a sin(t), which traces a circle of radius a as t varies. To find dy/dx for a parametric curve, find dy/dt and dx/dt separately, ordinary derivatives with respect to t, then divide: dy/dx = (dy/dt)/(dx/dt), valid wherever dx/dt is not zero. For the circle above, dy/dt = a cos(t) and dx/dt = -a sin(t), giving dy/dx = -cos(t)/sin(t) = -cot(t). Finally, since the derivative of a function is itself a new function, it can be differentiated again. The derivative of the derivative is called the second derivative, written f double-prime(x) or d squared y/dx squared, found simply by applying the same differentiation rules a second time to f prime(x). For y = x cubed, the first derivative is 3x squared, and differentiating that again gives the second derivative, 6x. The second derivative will matter enormously in the very next chapter of this thread, where it becomes the tool for telling a hill-shaped maximum point apart from a valley-shaped minimum point on a graph, without needing to look at the picture at all.