Calculus, the Second Time Through: Two Ideas, a Mirror, and Everything Else

The last article built two moves from scratch: shrink a secant line down to a tangent (the derivative), and shrink rectangles down to an exact area (the integral). Both worked perfectly for x², because x² is simple enough to expand by hand. Real problems rarely hand you x². They hand you sin(x)·e^x, or 1/(x²−1), or an equation that only tells you how fast something is changing and asks you to reconstruct what it actually is.

This article is about those two original moves surviving contact with everything else. Every rule below isn't a separate thing to memorize, it's the secant-to-tangent idea, or the rectangle-to-area idea, worked out one more time against a harder target. Watch for that pattern. It's the same pattern every time.

The Product Rule: Watching a Rectangle Grow

Two quantities, u and v, multiplied together, is naturally a rectangle: one side has length u, the other has length v, and the area is u·v. If both sides are changing (both u and v are functions of x), how fast does the area change?

Picture the rectangle growing outward on two edges at once, the right edge stretches out by a small amount $\Delta u$, and the top edge stretches up by a small amount $\Delta v$. The new, bigger area splits cleanly into four pieces:

Total new area $= uv + v\Delta u + u\Delta v + \Delta u\Delta v$. Subtract the original area to get just the change: $\Delta A = v\Delta u + u\Delta v + \Delta u\Delta v$. Divide everything by $\Delta x$ and let $\Delta x$ shrink toward zero, exactly the same shrinking move from the last article. Picture the two strips settling into ordinary numbers as $du/dx$ and $dv/dx$, while the tiny corner region, growing in both directions simultaneously, shrinks to nothing faster than either strip does, and drops out of the limit entirely. What's left:

Worked example: differentiate $x^2 \sin(x)$. Let $u = x^2$ (so $du/dx = 2x$) and $v = \sin(x)$ (so $dv/dx = \cos(x)$, derived two sections from now). Then:

$$\dfrac{d}{dx}(x^2 \sin x) = x^2 \cos(x) + \sin(x) \cdot 2x$$

No expanding, no first-principles limit needed ever again for a product, just the two pieces' own rates, cross-weighted.

The Chain Rule: Rates Inside Rates

Some quantities depend on something that itself depends on something else. A balloon's volume depends on its radius. Its radius, while someone's inflating it, depends on time. If you want to know how fast the volume is growing per second, and you only know how volume depends on radius, and how radius depends on time, the chain rule is the tool that connects the two.

Worked example, the balloon: volume $V = \frac{4}{3}\pi r^3$, and the radius is currently 5 cm, growing at $dr/dt = 2$ cm per second. How fast is the volume growing right now?

$$\dfrac{dV}{dr} = 4\pi r^2 \quad \text{(power rule, applied to the } r^3 \text{ term)}$$

At $r = 5$: $\dfrac{dV}{dr} = 4\pi(5)^2 = 100\pi$

$$\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt} = 100\pi \times 2 = 200\pi \approx 628.3 \text{ cubic centimeters per second}$$

The balloon shows the chain rule doing real-world work, connecting two separate, physically meaningful rates. It's worth also seeing the more everyday version, differentiating a single function that simply has another function tucked inside it. Take $\sin(3x^2+1)$. The "outer" function is sine, the "inner" function is $3x^2+1$:

$$\dfrac{d}{dx}\sin(3x^2+1) = \cos(3x^2+1) \times \dfrac{d}{dx}(3x^2+1) = \cos(3x^2+1) \times 6x = 6x\cos(3x^2+1)$$

Same rule, same multiplication, just with the "outer rate" and "inner rate" both living inside a single formula instead of two separate real-world quantities. That's the entire chain rule: multiply the two rates that sit between the thing you care about and the thing you actually control. Every "rate of a rate" problem, a shadow lengthening as someone walks away from a lamppost, a ripple's radius growing as its area grows, temperature changing as altitude changes as a plane climbs, is this same multiplication, just with different letters.

The Quotient Rule: The Product Rule in Disguise

A quotient $u/v$ is really $u$ times $v^{-1}$, so the quotient rule isn't a new idea at all, it's the product rule combined with the power rule applied to a negative exponent.

Worked example: derive the derivative of $\tan(x)$, a result that isn't obvious on its own and can't be found by simplifying first. Since $\tan(x) = \sin(x)/\cos(x)$, let $u=\sin x$ ($du/dx=\cos x$) and $v=\cos x$ ($dv/dx=-\sin x$):

$$\dfrac{d}{dx}[\tan x] = \dfrac{\cos x(\cos x) - \sin x(-\sin x)}{\cos^2 x} = \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x} = \dfrac{1}{\cos^2 x} = \sec^2 x$$

using the Pythagorean identity $\cos^2x+\sin^2x=1$ to collapse the numerator into something clean. That's a genuinely new result, the derivative of tangent, arrived at with nothing but the quotient rule and one trig identity.

Worked check with numbers, on a case simple enough to verify independently: differentiate $x^2/x$ at the point $x=3$, where the honest answer should just be the derivative of $x$ itself, which is 1. Let $u=x^2$ ($du/dx=2x$), $v=x$ ($dv/dx=1$).

$$\dfrac{d}{dx}\left(\dfrac{x^2}{x}\right) = \dfrac{x(2x) - x^2(1)}{x^2} = \dfrac{2x^2 - x^2}{x^2} = \dfrac{x^2}{x^2} = 1$$

Confirms the rule against something already checkable by simplifying first.

Derivatives of Sine and Cosine: A Circle Doing the Proving

Here's a genuinely different kind of function, $\sin(x)$, and the shrinking-interval definition from the last article still applies directly:

$$f'(x) = \lim_{h \to 0} \dfrac{\sin(x+h) - \sin(x)}{h}$$

Using the angle-addition identity $\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)$, this splits into two separate pieces:

$$f'(x) = \sin(x) \times \lim_{h\to 0}\dfrac{\cos h - 1}{h} \;+\; \cos(x) \times \lim_{h\to 0}\dfrac{\sin h}{h}$$

Both of those small limits get answered by the same picture, the unit circle: mark a small angle $\theta$ from the positive x-axis to a point P on the circle, drop a vertical line from P down to the x-axis (that segment's length is exactly $\sin\theta$), and draw the circular arc from $(1,0)$ to P (that arc's length is exactly $\theta$). For a small angle, that vertical segment and that arc are already nearly the same length, and they converge completely as $\theta$ shrinks toward zero.

Made rigorous rather than just eyeballed, comparing $\sin\theta$ against $\theta$ against a third quantity, $\tan\theta$ (the length you'd get by extending the radius line out to the vertical tangent at $(1,0)$), gives the inequality $\sin\theta < \theta < \tan\theta$ for any small positive angle. Divide through by $\sin\theta$: $1 < \theta/\sin\theta < 1/\cos\theta$. As $\theta$ shrinks toward zero, $\cos\theta$ approaches 1, so $1/\cos\theta$ approaches 1 too, squeezing $\theta/\sin\theta$ between two things that both approach 1. It has no choice but to approach 1 as well, which means $\lim_{h\to 0}\sin(h)/h = 1$. The second limit, $\lim_{h\to 0}(\cos h - 1)/h$, works out to 0 through a similar squeeze argument.

Substitute both limit values back into the split expression from the start of this section:

$$f'(x) = \sin(x) \times 0 \;+\; \cos(x) \times 1 = \cos(x)$$

Derivatives of Exponentials and Logarithms: Why e Is the "Natural" Base

Apply the shrinking-interval definition to $e^x$ directly:

$$\dfrac{d}{dx}[e^x] = \lim_{h \to 0} \dfrac{e^{x+h} - e^x}{h} = e^x \times \lim_{h \to 0} \dfrac{e^h - 1}{h}$$

That remaining limit, $(e^h - 1)/h$ as $h$ shrinks to zero, isn't a coincidence to memorize, it's essentially the definition of the number $e$: $e$ is the unique base for which this particular limit equals exactly 1.

Substitute that value directly back into the expression above:

$$\dfrac{d}{dx}[e^x] = e^x \times 1 = e^x$$

Logarithms fall out of this almost for free, using implicit differentiation, properly introduced in the next section, but here's the preview. If $y = \ln(x)$, then by definition $x = e^y$. Differentiate both sides with respect to $x$, remembering that $y$ is secretly a function of $x$:

$$1 = e^y \times \dfrac{dy}{dx}$$

Since $e^y = x$ by the original definition, substitute it back in:

$$1 = x \times \dfrac{dy}{dx} \quad \Rightarrow \quad \dfrac{dy}{dx} = \dfrac{1}{x}$$

Implicit Differentiation: When y Refuses to Be Isolated

Every function differentiated so far has had $y$ sitting by itself on one side: $y = x^2$, $y = \sin(x)$. Plenty of genuinely useful equations don't cooperate. A circle of radius 5 is written $x^2 + y^2 = 25$, and there's no way to isolate $y$ as a single clean function, the top half and bottom half are two different functions glued together at the edges.

The trick: differentiate both sides with respect to $x$ anyway, treating $y$ as an unnamed function of $x$, and every time $y$ shows up, the chain rule quietly kicks in, contributing an extra $dy/dx$.

$$x^2 + y^2 = 25$$ $$\dfrac{d}{dx}[x^2] + \dfrac{d}{dx}[y^2] = \dfrac{d}{dx}[25]$$ $$2x + 2y\dfrac{dy}{dx} = 0$$ $$\dfrac{dy}{dx} = -\dfrac{x}{y}$$

Worked check at the point $(3, 4)$, which sits on this circle since $3^2 + 4^2 = 9 + 16 = 25$:

$$\dfrac{dy}{dx} = -\dfrac{3}{4}$$

That claim survives a direct geometric test rather than sitting as an unverified formula. The radius from the origin to $(3, 4)$ has slope $4/3$. The tangent line at that same point, drawn from the implicit derivative, has slope $-3/4$. A tangent line to a circle is always perpendicular to the radius at that point, and the negative reciprocal of $4/3$ is exactly $-3/4$, so the two lines agree completely. The implicit derivative and a basic geometry fact confirm each other, which is a good sign the technique is doing something real, not just producing symbols.

u-Substitution: Undoing the Chain Rule

Every technique so far has been about taking derivatives apart. Integration is the reverse direction, and it turns out every differentiation rule has a mirror-image integration technique built by literally running it backward.

The chain rule says $\dfrac{d}{dx}[\sin(x^2)] = \cos(x^2) \times 2x$. Flip that statement around: if you're ever handed the integral $\int 2x\cos(x^2)\,dx$, and you notice that $2x$ is exactly the derivative of the $x^2$ sitting inside the cosine, you're looking at a chain rule in reverse, and the antiderivative is just $\sin(x^2)$.

The formal bookkeeping for this is called u-substitution:

Worked check: differentiate the answer. $\dfrac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot 2x$, exactly the original integrand. u-substitution isn't a new idea, it's pattern-matching for "does this integral secretly contain a chain rule that already happened," then running it in reverse.

Everything above handles an indefinite integral, no boundaries, just an antiderivative plus a constant. Definite integrals need one extra piece of bookkeeping: when $x$ is replaced by $u$, the boundary values need to switch to matching $u$-values too.

Worked example: $\displaystyle\int_0^2 2x\cos(x^2)\,dx$. Using the same substitution as above, $u=x^2$, the boundaries transform along with everything else: when $x=0$, $u=0^2=0$; when $x=2$, $u=2^2=4$.

$$\int_0^2 2x\cos(x^2)\,dx = \int_0^4 \cos(u)\,du = \big[\sin(u)\big]_0^4 = \sin(4) - \sin(0) = \sin(4)$$

Worked check: find the antiderivative in terms of $x$ first (already done above, $\sin(x^2)$), then evaluate directly at the original $x$-boundaries instead: $\sin(2^2) - \sin(0^2) = \sin(4) - \sin(0) = \sin(4)$. Both routes land on the identical answer, which is the actual point: changing the boundaries during substitution isn't a separate rule to memorize, it's just doing the "substitute back to $x$" step early, once, instead of saving it for the very end.

Integration by Parts: Undoing the Product Rule, and Why the Trade Is Worth Making

The product rule, rearranged, gives a genuinely useful integration formula. Starting from $\dfrac{d}{dx}(uv) = u'v + uv'$, integrate both sides:

$$uv = \int u'v\,dx + \int uv'\,dx$$

Rearranged into its standard, more usable form:

Here's the actual intuition behind why anyone would want to make this trade, since the formula alone doesn't explain it. Every integration-by-parts problem hands you a product of two pieces, and everything depends on which piece becomes $u$ and which becomes $dv$. The piece called $u$ should be whichever one gets simpler when differentiated, $x$ differentiates down to just 1, $\ln(x)$ differentiates down to $1/x$. The piece called $dv$ should be whichever one stays perfectly manageable when integrated, $e^x$ integrates right back to $e^x$, $\sin(x)$ integrates to $-\cos(x)$. The trade deliberately routes the "get simpler" job to whichever piece is actually capable of getting simpler, and routes the "stay easy" job to whichever piece can absorb it without complaint.

This is also where the classic ordering, logarithmic, inverse trig, algebraic, trigonometric, exponential (sometimes taught as the acronym LIATE), actually comes from, not as an arbitrary sequence to memorize, but as a genuine ranking of which function types simplify fastest under differentiation. A logarithm differentiates into something dramatically simpler. An exponential differentiates into itself, no simplification at all. So when a product contains both, the logarithm becomes $u$, since it has somewhere useful to go, and the exponential becomes $dv$, since it doesn't mind being integrated.

Worked example: $\int x e^x\,dx$. Here $x$ is algebraic (simplifies to 1 under differentiation) and $e^x$ is exponential (stays exactly itself either way), so $u = x$ ($du = dx$) and $dv = e^x\,dx$ ($v = e^x$):

$$\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C$$

Worked check: differentiate $xe^x - e^x$. Using the product rule on the first term: $e^x + xe^x - e^x = xe^x$, exactly the original integrand.

Sometimes neither piece simplifies in either direction, they just trade places with each other forever, and integration by parts still gets there, by a stranger route worth seeing once.

Worked example: $\int e^x \cos(x)\,dx$. Apply integration by parts, letting $u = e^x$ and $dv = \cos(x)\,dx$, so $du = e^x\,dx$ and $v = \sin x$:

$$\int e^x \cos x\,dx = e^x \sin x - \int e^x \sin x\,dx$$

The remaining integral needs integration by parts too. Apply it again, letting $u = e^x$ and $dv = \sin(x)\,dx$, so $v = -\cos x$:

$$\int e^x \sin x\,dx = -e^x \cos x + \int e^x \cos x\,dx$$

Substitute this back into the first line:

$$\int e^x \cos x\,dx = e^x \sin x - \left[-e^x \cos x + \int e^x \cos x\,dx\right] = e^x \sin x + e^x \cos x - \int e^x \cos x\,dx$$

The original integral has quietly reappeared on the right-hand side. Treat it like an unknown to solve for, calling it $I$:

$$I = e^x \sin x + e^x \cos x - I$$ $$2I = e^x \sin x + e^x \cos x$$ $$I = \dfrac{e^x \sin x + e^x \cos x}{2} + C$$

Worked check: differentiate $(e^x \sin x + e^x \cos x)/2$, using the product rule on each term: $[e^x\sin x + e^x\cos x + e^x\cos x - e^x\sin x]/2 = [2e^x\cos x]/2 = e^x\cos x$, exactly the original integrand. Integration by parts didn't just simplify this integral, it looped back on itself, and the loop itself was the solution, treating the unsolved integral as an algebraic unknown and solving for it directly.

Integration of Special Forms

Trig integrals using identities and substitution. The right technique for a trig integral depends on a single question: is the power even or odd?

For an even power, like $\int \sin^2(x)\,dx$, there's no obvious antiderivative sitting in plain sight, $\sin^2(x)$ isn't the derivative of anything simple. The half-angle identity $\sin^2(x) = \dfrac{1 - \cos 2x}{2}$ rewrites it into something that is:

$$\int \sin^2(x)\,dx = \int \dfrac{1 - \cos 2x}{2}\,dx = \dfrac{x}{2} - \dfrac{\sin(2x)}{4} + C$$

Worked check: differentiate $x/2 - \sin(2x)/4$. Using the chain rule on the second term: $1/2 - (2\cos 2x)/4 = 1/2 - \cos(2x)/2 = (1-\cos 2x)/2 = \sin^2(x)$, matching the same identity used to set the problem up.

For an odd power, a genuinely different move works better, worth seeing since it's a different trick entirely. $\int \sin^3(x)\,dx$: peel off a single factor of $\sin(x)$, leaving an even power behind, then rewrite that even power using $\sin^2(x) = 1 - \cos^2(x)$:

$$\int \sin^3(x)\,dx = \int \sin^2(x)\sin(x)\,dx = \int (1-\cos^2 x)\sin(x)\,dx$$

The integral is now secretly a u-substitution: let $u = \cos(x)$, so $du = -\sin(x)\,dx$, meaning $\sin(x)\,dx = -du$:

$$= \int (1-u^2)(-du) = -\int(1-u^2)\,du = -u + \dfrac{u^3}{3} + C = -\cos(x) + \dfrac{\cos^3(x)}{3} + C$$

Worked check: differentiate $-\cos x + \cos^3 x/3$. Using the chain rule on the second term: $\sin x + 3\cos^2 x(-\sin x)/3 = \sin x - \cos^2 x\sin x = \sin x(1-\cos^2 x) = \sin x \sin^2 x = \sin^3 x$, matching the original integrand.

The decision rule underneath both examples: an even power needs an identity to break it into pieces with a direct antiderivative; an odd power can always spare one factor to feed a substitution, since peeling off one sine leaves an even power that converts cleanly into cosines.

Rational functions using partial fractions. $\int \dfrac{1}{x^2-1}\,dx$ looks intractable until the denominator factors: $x^2 - 1 = (x-1)(x+1)$. The technique splits the single fraction into two simpler ones:

$$\dfrac{1}{(x-1)(x+1)} = \dfrac{A}{x-1} + \dfrac{B}{x+1}$$

$\dfrac{1}{x^2-1} = \dfrac{1/2}{x-1} - \dfrac{1/2}{x+1}$, and each piece integrates directly to a logarithm:

$$\int \dfrac{1}{x^2-1}\,dx = \dfrac{1}{2}\ln|x-1| - \dfrac{1}{2}\ln|x+1| + C$$

It's worth seeing why this technique works at all, rather than trusting the algebra on faith. Adding two simple fractions over a common denominator, $1/(x-1)$ plus $1/(x+1)$, produces exactly $[(x+1)+(x-1)] / [(x-1)(x+1)] = 2x/(x^2-1)$, a single, messier-looking fraction. Partial fractions is that exact process, run in reverse: given the messy combined fraction, find the two simple pieces that would have combined to produce it. Two separate $1/(\text{linear})$ terms are already known how to integrate. The technique's entire value is turning one unfamiliar integral into two familiar ones, using nothing but the algebra of adding fractions, done backward.

Differential Equations, Revisited: The Integrating Factor

The last article solved $dy/dt = ky$ by separating variables, but plenty of differential equations don't separate cleanly. The general first-order linear form is $dy/dx + P(x)y = Q(x)$, and the fix is a genuinely clever trick, one worth deriving rather than just accepting as a formula.

The goal: find some function $\mu(x)$ to multiply the whole equation by, chosen specifically so the left-hand side collapses into the derivative of a single product, $\mu \cdot y$. Start from what that would actually require. The product rule says:

$$\dfrac{d}{dx}[\mu \cdot y] = \mu\dfrac{dy}{dx} + y\dfrac{d\mu}{dx}$$

Compare that to what's sitting on the left side of the original equation after multiplying through by $\mu$: $\mu\dfrac{dy}{dx} + \mu Py$. For these two expressions to match exactly, the second terms have to be equal:

$$y\dfrac{d\mu}{dx} = \mu P y \quad \Rightarrow \quad \dfrac{d\mu}{dx} = \mu P$$

This is itself a differential equation for $\mu$, and it's the exact same separable equation solved earlier in this article, just with different letters. Separate it:

$$\dfrac{d\mu}{\mu} = P\,dx \quad \Rightarrow \quad \ln(\mu) = \int P\,dx \quad \Rightarrow \quad \mu = e^{\int P\,dx}$$

Worked example, a bathtub filling while draining, a genuinely common shape for this kind of equation: $dy/dx + 2y = 10$. Here $P(x) = 2$, a constant, so:

$$\mu(x) = e^{\int 2\,dx} = e^{2x}$$

Multiply the whole equation by $e^{2x}$:

$$e^{2x}\dfrac{dy}{dx} + 2e^{2x}y = 10e^{2x}$$

The entire left side is, by design, exactly $\dfrac{d}{dx}[e^{2x}y]$. So:

$$\dfrac{d}{dx}[e^{2x}y] = 10e^{2x}$$

Integrate both sides directly:

$$e^{2x}y = 5e^{2x} + C \quad \Rightarrow \quad y = 5 + Ce^{-2x}$$

Worked check: $dy/dx = -2Ce^{-2x}$. Plug back into the original equation: $-2Ce^{-2x} + 2(5+Ce^{-2x}) = -2Ce^{-2x}+10+2Ce^{-2x} = 10$. And there's a genuinely nice physical reading here: as $x$ grows large, the $Ce^{-2x}$ term fades to nothing, and $y$ settles toward 5, a steady equilibrium.

That equilibrium claim holds regardless of the starting point. Three solution curves, starting from three different values ($C = -3$, $C = 3$, and $C = 6$, corresponding to starting values of 2, 8, and 11), all bend toward the same horizontal line at $y = 5$ as $x$ grows, no matter where each one began. This single differential equation doesn't have one solution, it has an entire family of them, one curve for every possible starting value, and the starting condition only ever changes $C$, never where the family eventually ends up.

Slope Fields: Seeing a Solution Before Solving Anything

Every differential equation solved so far needed algebra to actually produce an answer. There's a way to see what the family of solutions looks like without doing any of that algebra at all.

A differential equation like $dy/dx = f(x, y)$ hands you, at every single point in the plane, the exact slope a solution curve would have if it happened to pass through that point. Plot a short line segment with that slope at a grid of points, and the resulting picture, called a slope field, lets a solution curve's shape emerge visually, just by imagining a curve that stays tangent to the little segments as it moves through the grid.

Take the equation $dy/dx = -x/y$, which happens to be exactly the implicit-differentiation relationship derived earlier for the circle $x^2+y^2=25$. Plotting short slope-segments at a grid of points, each tilted according to the local value of $-x/y$, visibly curls into circular arcs, matching a reference circle $x^2+y^2=2$ laid over the same grid, all without separating a single variable or integrating anything.

That the slope field for $dy/dx = -x/y$ visually traces circles isn't a coincidence, it's the same relationship found earlier by implicitly differentiating $x^2+y^2=C$. Solving that differential equation algebraically (separating variables: $y\,dy = -x\,dx$, integrating both sides: $y^2/2 = -x^2/2 + \text{constant}$, rearranging: $x^2+y^2 = \text{constant}$) produces exactly the family of circles the slope field already showed by eye. Seeing the answer first, then confirming it with algebra, is often a faster and more honest way to understand what a differential equation is actually describing.

How These Techniques Actually Relate to Each Other

Nothing in this article is a separate trick. Every integration technique is a differentiation rule, run in reverse, on purpose.

Conclusion & Key Takeaways

Nothing in this article introduced a new idea. The secant collapsing into a tangent, and the rectangles collapsing into an exact area, from the last article, are still the entire engine. Every rule here is what happens when that engine gets pointed at a harder target: two functions multiplied together, one function nested inside another, a circle instead of a simple curve, an equation that only describes a rate instead of handing you the answer directly. Learn the target, not a separate formula for each one, and the formulas stop needing to be memorized at all, they're just what the shrinking-interval trick was always going to produce, run one more time.

Sources

Paul Dawkins, Paul's Online Math Notes, Calculus I: derivative rules and implicit differentiation Paul Dawkins, Paul's Online Math Notes, Calculus II: integration by parts Paul Dawkins, Paul's Online Math Notes, Calculus II: integration techniques including partial fractions MIT OpenCourseWare 18.03SC, Differential Equations, Solutions of First Order Linear Equations MIT OpenCourseWare 18.03SC, Differential Equations, practice problems on first-order linear ODEs and direction fields James Stewart, Calculus, standard reference for derivative rules, integration techniques, and the unit circle derivation of the derivatives of sine and cosine; George B. Thomas, Thomas' Calculus, standard reference for implicit differentiation and partial fraction decomposition