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Why is the natural exponential function its own derivative, and how do we differentiate exponential functions in general?

Establish and use the derivative of the exponential function, including the chain rule for e raised to a function of x

WACE Year 12 Mathematics Methods Unit 3 derivatives of exponential functions: why e to the x is its own derivative, the chain rule for e to a function of x, base-a exponentials, and worked SCSA-style differentiation.

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  1. What this dot point is asking
  2. Why exe^{x} is its own derivative
  3. The chain rule with exponentials
  4. General bases
  5. Exponential models and proportional growth
  6. Reading the gradient

What this dot point is asking

SCSA Unit 3 singles out the natural exponential function exe^{x} because of one remarkable property: its gradient at every point equals its height. This dot point asks you to know that property, derive the chain-rule extension, and differentiate exponentials confidently in both the calculator-free and calculator-assumed sections of the WACE written examination.

Why exe^{x} is its own derivative

Consider f(x)=axf(x)=a^{x}. From first principles,

f(x)=limh0ax+haxh=axlimh0ah1h.f'(x)=\lim_{h\to 0}\frac{a^{x+h}-a^{x}}{h}=a^{x}\lim_{h\to 0}\frac{a^{h}-1}{h}.

The factor axa^{x} comes out because it does not depend on hh. The remaining limit is a constant that depends only on the base aa; it is the gradient of axa^{x} at x=0x=0. The number e2.71828e\approx 2.71828 is defined precisely as the base for which that limiting constant is exactly 11. Therefore
ddxex=ex.\frac{d}{dx}e^{x}=e^{x}.

This makes exe^{x} the natural model for any process whose rate of growth is proportional to its current size.

The chain rule with exponentials

Whenever the exponent is more than a single xx, the chain rule supplies a factor equal to the derivative of the exponent. For y=ekxy=e^{kx} the exponent is kxkx, whose derivative is kk, giving y=kekxy'=k\,e^{kx}. For a general inner function f(x)f(x) the factor is f(x)f'(x).

General bases

For a base other than ee, write ax=exlnaa^{x}=e^{x\ln a} and apply the chain rule. The exponent xlnax\ln a has derivative lna\ln a, so

ddxax=lnaexlna=axlna.\frac{d}{dx}a^{x}=\ln a\cdot e^{x\ln a}=a^{x}\ln a.

For example ddx2x=2xln2\dfrac{d}{dx}2^{x}=2^{x}\ln 2. When a=ea=e the factor lne=1\ln e=1 recovers ddxex=ex\dfrac{d}{dx}e^{x}=e^{x}.

Exponential models and proportional growth

The property ddxekx=kekx\dfrac{d}{dx}e^{kx}=k\,e^{kx} makes the exponential the natural model for any quantity whose rate of change is proportional to its current size. If Q=Q0ektQ=Q_{0}e^{kt}, then

dQdt=kQ0ekt=kQ,\frac{dQ}{dt}=kQ_{0}e^{kt}=kQ,

so the growth rate is always kk times the current amount. A positive kk gives growth (populations, compound interest); a negative kk gives decay (radioactive material, cooling, drug concentration). Recognising dQdt=kQ\dfrac{dQ}{dt}=kQ as the signature of exponential change is a recurring SCSA theme, and the constant kk is read straight off the exponent.

Reading the gradient

Because ex>0e^{x}>0 for all xx, the curve y=exy=e^{x} is always increasing and always concave up: both yy' and yy'' equal exe^{x}, which is positive everywhere. This is why the graph has no stationary points and a horizontal asymptote y=0y=0 as xx\to-\infty. The same positivity means y=ekxy=e^{kx} never has a stationary point either: kekx=0k e^{kx}=0 has no solution, so exponential curves are strictly monotonic.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20214 marksCalculator-free. Differentiate each function with respect to xx: (a) y=e4xy=e^{4x}; (b) y=x2e3xy=x^{2}e^{-3x}; (c) y=esinxy=e^{\sin x}.
Show worked answer →

Calculator-free differentiation across three rule types.

(a) Chain rule on the exponent 4x4x: dydx=4e4x\dfrac{dy}{dx}=4e^{4x}.

(b) Product rule with u=x2u=x^{2}, v=e3xv=e^{-3x}. Here u=2xu'=2x and v=3e3xv'=-3e^{-3x}, so dydx=2xe3x+x2(3e3x)=e3x(2x3x2)=xe3x(23x)\dfrac{dy}{dx}=2xe^{-3x}+x^{2}(-3e^{-3x})=e^{-3x}(2x-3x^{2})=xe^{-3x}(2-3x).

(c) Chain rule with inner f(x)=sinxf(x)=\sin x, f(x)=cosxf'(x)=\cos x: dydx=cosxesinx\dfrac{dy}{dx}=\cos x\,e^{\sin x}.

Markers reward the chain-rule factor in (a) and (c), the product rule with the correct internal chain rule in (b), and a tidy factored form.

WACE 20236 marksCalculator-assumed. The number of bacteria is modelled by N(t)=500e0.18tN(t)=500e^{0.18t}, where tt is in hours. (a) Find the rate of growth N(t)N'(t). (b) Show that the rate of growth is proportional to NN and state the constant of proportionality. (c) Find the rate of growth, to the nearest whole number, when t=5t=5.
Show worked answer →

A 6 mark applied differentiation question.

(a) Differentiate using the chain rule on the exponent 0.18t0.18t: N(t)=500×0.18e0.18t=90e0.18tN'(t)=500\times 0.18\,e^{0.18t}=90e^{0.18t} bacteria per hour.

(b) Factor to compare with NN: N(t)=90e0.18t=0.18×500e0.18t=0.18N(t)N'(t)=90e^{0.18t}=0.18\times 500e^{0.18t}=0.18\,N(t). So N(t)=0.18NN'(t)=0.18N, proportional to NN with constant of proportionality 0.180.18 per hour. This is the defining feature of exponential growth.

(c) At t=5t=5: N(5)=90e0.9=90×2.4596221N'(5)=90e^{0.9}=90\times 2.4596\approx 221 bacteria per hour.

Markers reward the chain-rule derivative, expressing NN' as a multiple of NN, and the correct evaluated rate with units.

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