🌀Transcendental Derivatives

Every function you have differentiated so far has been built out of powers of x, and the power rule handled all of them. The functions in this craft are different: they are transcendental, which means no amount of algebra with powers of x will produce them. Exponentials, logarithms and the tangent family all sit outside the reach of the power rule, and each one needs a rule of its own.

Start with the exponential, and with the reason e exists at all. Look at what happens to 2^x when x moves a tiny step h:

(2^x+h - 2^x)/h = 2^x * (2^h - 1)/h

The 2^x comes straight out of the front, so the derivative of 2^x is 2^x multiplied by some fixed number — the limit of (2^h - 1)/h. For base 2 that number is about 0.693. For base 3 it is about 1.099. Somewhere between 2 and 3 there is a base where the number is exactly 1, and that base is e. So:

d/dx of e^x = e^x

e^x is the one function that is its own derivative. That is the whole reason it is everywhere in this subject.

Any other base goes through e. Since b = e^{ln b}, you can write b^x = e^x*ln(b), and the chain rule then gives

d/dx of b^x = ln(b)*b^x

The fixed number that appeared above is ln(b): ln(2) = 0.693..., ln(3) = 1.099..., and ln(e) = 1, which is why base e is the tidy one.

The exponent is where the chain rule enters. If the exponent is kx rather than x, the inside has derivative k, so the answer picks up a factor of k:

d/dx of e^3x = 3e^3x d/dx of 5e^2x = 10e^2x d/dx of 2^4x = 4*ln(2)*2^4x

The x is in the exponent, never in the base, so the power rule has nothing to bring down. Writing d/dx of 2^x as x*2^x-1 is the single most common slip in this whole craft.

The natural logarithm is the inverse of e^x, and its derivative comes straight out of that. Write y = ln(x), so e^y = x, and differentiate both sides with respect to x:

e^y * (dy/dx) = 1, so dy/dx = 1/e^y = 1/x

d/dx of ln(x) = 1/x (for x > 0)

With the chain rule attached, the rule becomes the one you will use every time:

d/dx of ln(u) = u'/u

the derivative of the inside, over the inside. Three consequences are worth knowing by heart.

• ◆ln(5x) has derivative 1/x, not 5/x. The chain rule gives 5/(5x), and the 5 cancels. The log law says the same thing: ln(5x) = ln(5) + ln(x), and ln(5) is a constant, so it contributes nothing. Multiplying the inside by a constant slides the graph up; it does not tilt it.
• ◆ln(x⁴) has derivative 4/x. Bring the exponent down first with the log law: ln(x⁴) = 4ln(x).
• ◆ln(2x + 3) has derivative 2/(2x + 3). Here the inside really does have a derivative worth keeping.

A logarithm to another base changes base first. Since log base b of x = ln(x)/ln(b), and ln(b) is a fixed number,

d/dx of (log base b of x) = 1/(x*ln(b))

Only base e gives the clean 1/x. Base 10 leaves 1/(x*ln(10)), and there is no way to make that ln(10) go away.

You already know that sin x differentiates to cos x and cos x to -sin x. Everything else in trigonometry is built from those two by division, so everything else comes from the quotient rule. Nothing here has to be memorised as a separate fact — you can rebuild any of the four in about fifteen seconds.

Tangent, for instance:

tan x = sin x / cos x (tan x)' = (cos x * cos x - sin x * (-sin x)) / cos² x = (cos² x + sin² x) / cos² x = 1 / cos² x = sec² x

Secant is the same move with u = 1 on top:

sec x = 1/cos x (sec x)' = (0 * cos x - 1 * (-sin x)) / cos² x = sin x / cos² x = (1/cos x)(sin x/cos x) = sec x tan x

The other two come out the same way, and the whole family is:

• ◆d/dx of tan x = sec² x
• ◆d/dx of sec x = sec x tan x
• ◆d/dx of cot x = -csc² x
• ◆d/dx of csc x = -csc x cot x

The two names that start with "co" are the two whose derivatives are negative. Cotangent and cosecant both decrease wherever they are defined, and their minus signs are part of the rule, not an optional extra.

When one of these has an inside, the chain rule attaches as usual: d/dx of tan(3x) = 3*sec²(3x), which written without the secant name is 3/cos(3x)². And when a slope is wanted at a specific angle, differentiate first and substitute second: at x = pi/3 the slope of y = tan x is sec²(pi/3) = 1/(1/2)² = 4.

Nothing new is needed once the individual rules are in place — the product, quotient and chain rules do not care which family a factor comes from. What changes is that you now have to notice which structure is outermost before you start.

Compositions. Differentiate the outer function, leave the inside exactly as it is, and multiply by the derivative of the inside.

d/dx of e^{x^2} = 2x*e^{x^2} d/dx of ln(x² + 1) = 2x/(x² + 1) d/dx of sin(e^x) = e^x*cos(e^x) d/dx of ln(cos x) = -sin(x)/cos(x)

Notice the third one: the outer sine is differentiated to a cosine of the same inside, e^x, not of x.

Products. Two terms, always.

d/dx of x*e^x = e^x + x*e^x = e^x*(x + 1) d/dx of x²*ln(x) = 2x*ln(x) + x = x*(2ln(x) + 1) d/dx of e^x*sin x = e^x*sin x + e^x*cos x = e^x*(sin x + cos x)

In the middle line, x² * (1/x) simplifies to x, which is why the answer is tidier than you might expect.

Quotients. The term with u' comes first, and the denominator is squared.

d/dx of (e^x/x) = (e^x*x - e^x)/x² = e^x*(x - 1)/x² d/dx of (ln(x)/x) = (x*(1/x) - ln x)/x² = (1 - ln x)/x²

A last piece of advice that saves more marks than any rule: factorise at the end, not in the middle. Get every term written down first, then look for what they share.

When x is in the base AND in the exponent, no rule applies. y = x^x is not a power (the exponent is not constant) and not an exponential (the base is not constant). The way through is to take a logarithm first, which drags the exponent down to where the product rule can reach it.

y = x^x ln y = x*ln x (take ln of both sides) (1/y)(dy/dx) = ln x + x*(1/x) = ln x + 1 (differentiate; the left side needs the chain rule) dy/dx = y*(ln x + 1) = x^x*(ln x + 1) (multiply back by y)

Take ln, differentiate, multiply back. The left-hand side always becomes (1/y)(dy/dx), because y is a function of x. Forgetting the last step — multiplying back by y — is what leaves an answer that is only half the size it should be.

The same trick tames an ugly product: for f(x) = x²*(x + 1)³, taking ln turns the product into a sum, ln f = 2ln x + 3ln(x + 1), so f'/f = 2/x + 3/(x + 1), and f' = f(x)*(2/x + 3/(x + 1)).

The inverse-function rule. The graph of an inverse g is the graph of f reflected in the line y = x. Reflecting swaps the run and the rise, so it turns a slope into its reciprocal — read at the mirrored point:

if f(a) = b, then g'(b) = 1/f'(a)

The point matters as much as the reciprocal. For f(x) = x³ + x, to find g'(2) you first solve f(a) = 2, which gives a = 1, and then take 1/f'(1) = 1/4. Using f'(2) instead would be reading the slope in the wrong place entirely.

That rule is where the inverse trigonometric derivatives come from. Writing y = arcsin x means sin y = x, so cos(y)*(dy/dx) = 1 and dy/dx = 1/cos y = 1/sqrt(1 - x²). The same argument on tan y = x gives 1/(1 + x²):

• ◆d/dx of arcsin x = 1/sqrt(1 - x²)
• ◆d/dx of arccos x = -1/sqrt(1 - x²)
• ◆d/dx of arctan x = 1/(1 + x²)

Two square roots and one without: the pair that came from sine and cosine share sqrt(1 - x²), and the tangent's has 1 + x² and no root at all.