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WAMath MethodsSyllabus dot point

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 $e^{x}$ is its own derivative
  3. The chain rule with exponentials
  4. General bases
  5. 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}.

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.