Differentiation

We covered the definition of differentiation and its basic properties in §Differentiation and Derivatives. In this post, we treat derivatives of concrete functions and differentiation rules applicable to general functions.

Term-by-Term Differentiation of Power Series

First, for any natural number \(n\), one can verify that the derivative of \(x^n\) at a point \(a\) is given by

\[\lim_{h\rightarrow 0}\frac{(a+h)^n-a^n}{h}=\lim_{h\rightarrow 0}\frac{na^{n-1}h+\ldots}{h}=na^{n-1}\]

and similarly, putting the difference quotient over a common denominator for a negative integer \(n = -m\) gives

\[\frac{(a+h)^{-m} - a^{-m}}{h} = -\frac{(a+h)^m - a^m}{h}\cdot\frac{1}{(a+h)^ma^m}\]

so ultimately \((x^n)' = n x^{n-1}\) holds for all integers \(n\).

On the other hand, by linearity a polynomial is differentiated term by term, so this also yields a differentiation rule for arbitrary polynomial functions. However, in general distributing differentiation term by term in an infinite sum is not obvious, and the following proposition guarantees that this is permitted for power series.

Proposition 1 (Term-by-term differentiation of power series) If the power series \(f(x) = \sum_{n=0}^\infty c_n x^n\) has radius of convergence \(R > 0\), then \(f\) is differentiable for \(\lvert x\rvert < R\) and

\[f'(x) = \sum_{n=1}^\infty n c_n x^{n-1}\]

and this series also has radius of convergence \(R\).

That the series on the right-hand side has radius of convergence \(R\) itself follows from the fact that as \(n\rightarrow \infty\), \(n^{1/n}\) converges to \(1\), but we lack the tools to rigorously justify that it actually equals \(f'(x)\), so we shall accept this and move on.

Derivatives of Trigonometric and Exponential Functions

Before examining differentiation rules in earnest, we derive the derivatives of various functions.

First, the exponential function was defined in §Power Series, ⁋Example 3 (Exponential function) by \(e^x = \sum_{n\geq 0} x^n/n!\). Applying Proposition 1 (Term-by-term differentiation of power series), its derivative is the sum of term-by-term derivatives

\[(e^x)' = \sum_{n=1}^\infty n \frac{x^{n-1}}{n!} = \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!} = \sum_{m=0}^\infty \frac{x^m}{m!} = e^x\]

so the exponential function is invariant under differentiation. In high school, it was from the limit

\[\lim_{h\rightarrow 0}(e^h-1)/h = 1\]

that this result was obtained.

Trigonometric functions will also soon be written in power-series form, but for now, using the two limits

\[\lim_{h \rightarrow 0} \frac{\sin h}{h} = 1, \qquad \lim_{h \rightarrow 0} \frac{1 - \cos h}{h} = 0\]

we obtain them. The first limit was obtained by the squeeze theorem in §Limits of Functions, ⁋Example 10, and the second limit follows from the inequality \(\lvert 1 - \cos h\rvert \leq h^2/2\) established in that same example, which yields \(\lvert(1 - \cos h)/h\rvert \leq \lvert h\rvert/2\).

Proposition 2 (Derivatives of trigonometric functions) At every point \((\sin x)' = \cos x\) and \((\cos x)' = -\sin x\).

Proof

Splitting the difference quotient using the addition formula \(\sin(x+h) = \sin x\cos h + \cos x \sin h\), we have

\[\frac{\sin(x+h) - \sin x}{h} = \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}\]

and as \(h \rightarrow 0\), by the two limits above this converges to \(\sin x \cdot 0 + \cos x \cdot 1 = \cos x\). The derivative of \(\cos x\) is obtained in the same way.

Various Differentiation Rules

We now define differentiation rules applicable to general forms. In §Differentiation and Derivatives, ⁋Proposition 4 we saw that differentiation behaves well with respect to constant multiples and addition, but it does not simply distribute over products.

Proposition 3 (Product rule) If \(f, g\) are differentiable at \(a\), then \(fg\) is also differentiable at \(a\) and

\[(fg)'(a) = f'(a) g(a) + f(a) g'(a)\]

holds.

Proof

Adding and subtracting the same term in the difference quotient gives

\[\frac{f(a+h)g(a+h) - f(a)g(a)}{h} = \frac{f(a+h)-f(a)}{h} g(a+h) + f(a) \frac{g(a+h)-g(a)}{h}\]

As \(h \rightarrow 0\), the difference quotient in the first term converges to \(f'(a)\), \(g(a+h)\) converges to \(g(a)\) by the continuity of \(g\) (§Differentiation and Derivatives, ⁋Proposition 2), and the difference quotient in the second term converges to \(g'(a)\), so by §Limits of Functions, ⁋Proposition 5 (Limit Laws) the sum converges to \(f'(a)g(a) + f(a)g'(a)\).

The most widely used rule is the derivative of a composite function.

Theorem 4 (Chain rule) If \(f\) is differentiable at \(a\) and \(g\) is differentiable at \(b = f(a)\), then the composition \(g \circ f\) is also differentiable at \(a\) and

\[(g \circ f)'(a) = g'(f(a)) f'(a)\]

holds.

Proof

We transfer the differentiability of \(g\) to the auxiliary function

\[\varphi(y) = \begin{cases} \frac{g(y) - g(b)}{y - b}, & y \neq b,\\[1mm] g'(b), & y = b \end{cases}\]

By definition of differentiability, \(\varphi\) is continuous at \(b\), and \(g(y) - g(b) = \varphi(y)(y - b)\) holds for all \(y\). Substituting \(y = f(a+h)\) gives

\[\frac{g(f(a+h)) - g(f(a))}{h} = \varphi(f(a+h)) \frac{f(a+h) - f(a)}{h}\]

and as \(h \rightarrow 0\), by the continuity of \(f\) we have \(\varphi(f(a+h)) \rightarrow \varphi(b) = g'(b)\), and the second factor converges to \(f'(a)\), so the limit is \(g'(f(a))f'(a)\).

The quotient rule now follows as a corollary of the product rule and the chain rule.

Corollary 5 (Quotient rule) If \(f, g\) are differentiable at \(a\) and \(g(a) \neq 0\), then \(f/g\) is also differentiable at \(a\) and

\[\left(\frac{f}{g}\right)'(a) = \frac{f'(a) g(a) - f(a) g'(a)}{g(a)^2}\]

holds.

Proof

Defining the function \(h(t)=1/t\), we have \(1/g =h \circ g\), and since \(h'(t)=-1/t^2\) by the derivative of integer powers, by Theorem 4 (Chain rule) we get \((1/g)'(a) = -g(a)^{-2}g'(a)\). Now applying the product rule to \(f/g = f\cdot(1/g)\), we obtain

\[\left(\frac{f}{g}\right)'(a) = \frac{f'(a)}{g(a)} - \frac{f(a)g'(a)}{g(a)^2} = \frac{f'(a)g(a) - f(a)g'(a)}{g(a)^2}\]

For instance, from \(\tan x = \sin x/\cos x\) we have \((\tan x)' = (\cos^2 x + \sin^2 x)/\cos^2 x = \sec^2 x\), and by the same method \((\cot x)' = -\csc^2 x\).

Finally, if a function is monotone, the derivative of its inverse is also obtained immediately without separate computation.

Proposition 6 (Derivative of an inverse function) Let \(f\) be a continuous monotone function on an interval having an inverse \(f^{-1}\), and suppose \(f\) is differentiable at \(a\) with \(f'(a) \neq 0\). Then \(f^{-1}\) is differentiable at \(b = f(a)\) and

\[(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{f'(f^{-1}(b))}\]

holds.

Proof

Let \(f\) be defined on the interval \(I\), and let its image be \(J = f(I)\). Since \(f\) is monotone and has an inverse, it is injective, hence strictly monotone, and therefore \(f^{-1} : J \rightarrow I\) is continuous by §Continuous Functions, ⁋Proposition 7.

Now, if \(y \in J\) satisfies \(y \neq b\), then \(x = f^{-1}(y)\) differs from \(a = f^{-1}(b)\) by injectivity, so we can rewrite the difference quotient of \(f^{-1}\) as

\[\frac{f^{-1}(y) - f^{-1}(b)}{y - b} = \frac{x - a}{f(x) - f(a)} = \left(\frac{f(x)-f(a)}{x-a}\right)^{-1}\]

As \(y \rightarrow b\), the continuity of \(f^{-1}\) gives \(x \rightarrow a\), and the expression in parentheses on the right converges to \(f'(a)\), so by \(f'(a) \neq 0\) and part 4 of §Limits of Functions, ⁋Proposition 5 (Limit Laws), the above difference quotient converges to \(1/f'(a)\). That is, \(f^{-1}\) is differentiable at \(b\) and \((f^{-1})'(b) = 1/f'(a)\).

With this, we can also proceed with differentiating the inverses of the functions examined above. For instance, for the inverse of \(e^x\), namely \(\ln\), since \((e^x)' = e^x\), we have \((\ln y)' = 1/e^{\ln y} = 1/y\), and for the inverse of \(\sin\) restricted to \((-\pi/2, \pi/2)\), namely \(\arcsin\), since \(f'(x) = \cos x = \sqrt{1 - \sin^2 x} > 0\), we have

\[(\arcsin y)' = \frac{1}{\sqrt{1 - y^2}} \qquad (\lvert y\rvert < 1)\]

and in the same manner, \((\arctan y)' = 1/(1 + y^2)\).

On the other hand, once we compute a derivative using the rules developed so far, we may ask whether that derivative is again continuous; yet even a function differentiable everywhere can have a discontinuous derivative.

Example 7 (A differentiable function that is not of class \(C^1\)) The function

\[f(x) = \begin{cases} x^2 \sin(1/x) & (x \neq 0) \\ 0 & (x = 0) \end{cases}\]

is differentiable at every point. For \(x \neq 0\), the product rule and chain rule give \(f'(x) = 2x\sin(1/x) - \cos(1/x)\), and at \(0\), the difference quotient is \(x\sin(1/x) \rightarrow 0\), so \(f'(0) = 0\). However, as \(x \rightarrow 0\), while \(2x\sin(1/x) \rightarrow 0\), \(\cos(1/x)\) oscillates infinitely often between \([-1, 1]\) and does not have a limit, so \(f'\) is discontinuous at \(0\). That is, \(f\) is differentiable everywhere, but because its derivative is not continuous, it is not of class \(C^1\) (in the sense of §Differentiation and Derivatives, ⁋Definition 5).

Nevertheless, a derivative cannot be discontinuous in just any manner. Even if a derivative is not continuous, it necessarily attains every intermediate value (Darboux’s theorem), so it cannot have a jump discontinuity, and the discontinuity exhibited in Example 7 (A differentiable function that is not of class \(C^1\)) is also due to oscillation rather than a jump.

Applications of Differentiation Rules

Finally, we conclude this post by examining how to apply the rules examined so far. First, although this direction somewhat puts the cart before the horse compared to the reason we introduced power series, Proposition 1 (Term-by-term differentiation of power series) can be used to evaluate series.

Example 8 (Sum of an infinite series) Let us differentiate both sides of the geometric series (§Infinite Series, ⁋Example 2):

\[\frac{1}{1-x} = \sum_{n=0}^\infty x^n \qquad (\lvert x\rvert < 1)\]

By the chain rule, the left-hand side is

\[\left(\frac{1}{1-x}\right)' = 1/(1-x)^2\]

and applying Proposition 1 (Term-by-term differentiation of power series) to the right-hand side gives \(\sum_{n\geq 1} n x^{n-1}\), so

\[\sum_{n=1}^\infty n x^{n-1} = \frac{1}{(1-x)^2} \qquad (\lvert x\rvert < 1)\]

and multiplying both sides by \(x\), we obtain

\[\sum_{n\geq 1} n x^n = \frac{x}{(1-x)^2}\]

For instance, substituting \(x = 1/2\), we can also compute the sum of the series \(\sum_{n\geq 1} n/2^n = 2\).

The second application is a result that fits this category of calculus a bit better, telling us a rule for actually differentiating a given function.

Example 9 (Various differentiation methods) When an equation defines \(y\) as a function of \(x\), finding an explicit expression of the form \(y=...\) in order to find \(y'\) is often inefficient, or taking such a form often does not yield a clean derivative. In this situation, viewing \(y\) as a function of \(x\), differentiating both sides, and then solving for \(y'\) is called implicit differentiation. For instance, differentiating both sides for the unit circle \(x^2 + y^2 = 1\) gives

\[2x + 2y y' = 0\]

so at points where \(y \neq 0\), we obtain \(y' = -x/y\), and thus can find the slope of the tangent line at such a point \((x_0,y_0)\).

For functions with a variable in the exponent, logarithmic differentiation (taking the logarithm of both sides and then differentiating) is convenient. For example, for \(y = x^x\) (\(x > 0\)), differentiating both sides of \(\ln y = x\ln x\) and using \((\ln y)' = y'/y\) together with the product rule gives

\[\frac{y'}{y} = \ln x + 1 \quad\Longrightarrow\quad y' = x^x(\ln x + 1)\]

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References

[Ste] J. Stewart, Calculus, 8th ed., Cengage Learning, 2016.
[Kim] 김홍종, 미적분학 1·2, 제3개정판, 서울대학교출판문화원, 2020.