Calculus I · A Toolkit for Slopes

The exponential, and the derivatives of inverse functions

The exponential cameo. One function deserves a special mention here and a full chapter (Chapter 5) of its own: $e^x$, the function that is its own derivative,

$$\frac{d}{dx}e^x = e^x.$$

(From the definition, the difference quotient is $e^x\cdot\tfrac{e^h - 1}{h}$, and the fact that $\tfrac{e^h-1}{h} \to 1$ is precisely what singles out the number $e$ — the loan you flagged in Chapter 3’s Looking-ahead, repaid in Chapter 5.) Combined with the chain rule, this differentiates every exponential you’ll meet: $\tfrac{d}{dx}e^{kx} = k\,e^{kx}$, $\tfrac{d}{dx}e^{x^2} = 2x\,e^{x^2}$.

Inverse functions. The last tool in the kit, and the one Chapter 5 needs to differentiate the logarithm. Suppose $g$ is the inverse of $f$ — meaning $f(g(x)) = x$ for all $x$. Differentiate both sides with the chain rule: the left side gives $f'(g(x))\cdot g'(x)$, the right side gives $1$. Solve:

Inverse-Function Rule. If $g = f^{-1}$, then $\displaystyle g'(x) = \frac{1}{f'\!\big(g(x)\big)}.$

The picture is clean: an inverse reflects a graph across the line $y = x$, which swaps “rise” and “run” — so the slope of the inverse at a point is the reciprocal of the slope of the original at the matching point. And it passes an honesty check against Chapter 3: $\sqrt{x}$ is the inverse of $f(x) = x^2$ (for $x \ge 0$), and since $f'(x) = 2x$, the rule gives $\tfrac{d}{dx}\sqrt{x} = \tfrac{1}{f'(\sqrt{x})} = \tfrac{1}{2\sqrt{x}}$ — the same answer you rationalized by hand, now obtained without a single difference quotient. This is exactly the machinery Chapter 5 will turn on $\ln x$ (the inverse of $e^x$) to discover that its derivative is $\tfrac1x$.

Toolkit: The rules of differentiation

Rule	Formula
Power	$\dfrac{d}{dx}x^n = n\,x^{n-1}$
Sum	$(f+g)' = f' + g'$
Constant multiple	$(cf)' = c\,f'$
Product	$(fg)' = f'g + fg'$
Quotient	$\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$
Chain	$\dfrac{d}{dx}f(g(x)) = f'(g(x))\,g'(x)$
Sine / Cosine	$(\sin x)' = \cos x,\quad (\cos x)' = -\sin x$
Exponential	$(e^x)' = e^x$
Inverse	$(f^{-1})'(x) = \dfrac{1}{f'(f^{-1}(x))}$

With these, you can differentiate essentially every function in this book — no difference quotients required. The art is recognizing which rule (or combination) a given function calls for.

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