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Pure · Differentiation

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Differentiating exponentials and logarithms

The standard results d/dx eˣ = eˣ, d/dx eᵏˣ = k eᵏˣ, d/dx aˣ = aˣ ln a and d/dx ln x = 1/x, why e is the special base, and combining them with the chain, product and quotient rules.

Pure · Differentiation

Differentiating exponentials and logarithms

The standard results d/dx eˣ = eˣ, d/dx eᵏˣ = k eᵏˣ, d/dx aˣ = aˣ ln a and d/dx ln x = 1/x, why e is the special base, and combining them with the chain, product and quotient rules.

Why it works

Differentiate any exponential y=axy = a^x from first principles and something striking happens. The gradient at xx works out to axa^x multiplied by a constant — the value of the gradient at x=0x = 0: ddxax=ax×(lim⁡h→0ah−1h).\frac{d}{dx}a^x = a^x \times \Big(\lim_{h\to 0}\frac{a^h - 1}{h}\Big). So the curve y=axy = a^x is, at every point, proportional to its own height. The only question is what that constant of proportionality is. For a=2a = 2 it is about 0.690.69; for a=3a = 3 it is about 1.101.10. Somewhere between 22 and 33 there is a base for which the constant is exactly 11 — and that base is the number e≈2.718e \approx 2.718. This is the whole reason ee is special: ddxex=ex.\boxed{\frac{d}{dx}e^x = e^x.} The exponential function exe^x is its own derivative. It is the function that grows at a rate equal to its current size — which is why it models unchecked growth and decay everywhere in science.

Scaling the exponent. For y=ekxy = e^{kx} the chain rule applies with inside u=kxu = kx: dydx=ekx×k=k ekx\frac{dy}{dx} = e^{kx}\times k = k\,e^{kx}. The constant kk drops out front. So ddxe3x=3e3x\frac{d}{dx}e^{3x} = 3e^{3x} and ddxe−x=−e−x\frac{d}{dx}e^{-x} = -e^{-x}.

Other bases. Writing a=eln⁡aa = e^{\ln a} gives ax=e(ln⁡a)xa^x = e^{(\ln a)x}, so by the same chain rule ddxax=axln⁡a.\frac{d}{dx}a^x = a^x \ln a. That mysterious constant from the start was simply ln⁡a\ln a all along (and ln⁡e=1\ln e = 1, recovering exe^x).

The logarithm. Since y=ln⁡xy = \ln x is the inverse of x=eyx = e^y, use the reciprocal connection: dxdy=ey=x\frac{dx}{dy} = e^y = x, so ddxln⁡x=1x.\frac{d}{dx}\ln x = \frac{1}{x}. A clean, memorable result — and the missing piece that lets you integrate 1x\frac1x. With the chain rule, ddxln⁡(f(x))=f′(x)f(x)\frac{d}{dx}\ln(f(x)) = \dfrac{f'(x)}{f(x)}: the derivative of the inside over the inside.

These four standard results combine with every rule you know. The exam favourites are xexx e^x and x2exx^2 e^x (product rule), exx\dfrac{e^x}{x} (quotient rule) and ln⁡(3x2+1)\ln(3x^2 + 1) (chain rule).