Learning module Derivatives · Lesson 3 of 5
Why the differentiation rules work
Derive the differentiation rules, then combine them in demanding calculations.
Allow 60–90 minutes, plus problem work.
Table of contents
Guiding question. Can every rule be traced back to the same limit, rather than memorized as an unrelated instruction?
Assume \(f\) and \(g\) are differentiable at \(a\). Write \(\Delta f=f(a+h)-f(a)\) and similarly for \(\Delta g\).
Linearity and the product rule
For constants \(\alpha,\beta\), the quotient for \(\alpha f+\beta g\) is \(\alpha\Delta f/h+\beta\Delta g/h\). Taking limits gives
\[(\alpha f+\beta g)'(a)=\alpha f'(a)+\beta g'(a).\]For a product, add and subtract \(f(a+h)g(a)\):
\[\frac{f(a+h)g(a+h)-f(a)g(a)}h =f(a+h)\frac{\Delta g}h+g(a)\frac{\Delta f}h.\]Differentiability gives continuity of \(f\), so \(f(a+h)\to f(a)\). Therefore
\[(fg)'(a)=f'(a)g(a)+f(a)g'(a).\]The two terms account for changes in each factor. For example, \((x\cdot x)'=2x\); multiplying the two derivatives would incorrectly give \(1\).
Reciprocals and quotients
If \(g(a)\ne0\), continuity makes \(g(a+h)\ne0\) for sufficiently small \(h\). Then
\[\frac{1/g(a+h)-1/g(a)}h =-\frac{\Delta g/h}{g(a+h)g(a)} \longrightarrow-\frac{g'(a)}{g(a)^2}.\]Apply the product rule to \(f\cdot(1/g)\):
\[\left(\frac fg\right)'(a) =\frac{f'(a)g(a)-f(a)g'(a)}{g(a)^2}.\]The condition \(g(a)\ne0\) matters. A formula derived on a punctured domain does not automatically define a derivative at a missing point.
The chain rule, including a subtle case
Suppose \(g\) is differentiable at \(a\) and \(f\) at \(b=g(a)\), with the composition defined near \(a\). Differentiability means we can write
\[g(a+h)-b=h\bigl(g'(a)+\varepsilon(h)\bigr),\qquad\varepsilon(h)\to0,\]and
\[f(b+k)-f(b)=k\bigl(f'(b)+\eta(k)\bigr),\qquad\eta(k)\to0.\]Define \(\eta(0)=0\) so the second identity also holds when \(k=0\). Substituting \(k=g(a+h)-b\), which tends to zero, gives
\[\frac{f(g(a+h))-f(g(a))}h =\bigl(g'(a)+\varepsilon(h)\bigr)\bigl(f'(b)+\eta(k)\bigr) \longrightarrow g'(a)f'(b).\]Thus \((f\circ g)'(a)=f'(g(a))g'(a)\). This proof also covers increments for which \(g(a+h)=g(a)\). Dividing by \(g(a+h)-g(a)\) without checking it could miss those increments.
A small toolbox, built from limits
For trigonometric functions, use radians and the standard limits
\[\frac{\sin h}h\to1,\qquad\frac{\cos h-1}h\to0.\]The angle-addition formulas turn the quotient for \(\sin x\) into
\[\sin a\frac{\cos h-1}h+\cos a\frac{\sin h}h\longrightarrow\cos a.\]The same method gives \((\cos x)'=-\sin x\). For the exponential, use the standard limit \((e^h-1)/h\to1\) and \(e^{a+h}=e^ae^h\) to obtain \((e^x)'=e^x\). These three elementary limits are inputs here; their proofs can be reviewed in a limits module. The logarithm and general real powers will be justified in Lesson 4.
| Function | Derivative | Domain for this formula |
|---|---|---|
| \(c\) | \(0\) | \(\mathbb R\) |
| \(x^n\), integer \(n\ge1\) | \(nx^{n-1}\) | \(\mathbb R\) |
| \(x^{-n}\), integer \(n\ge1\) | \(-nx^{-n-1}\) | \(x\ne0\) |
| \(\sqrt{x}\) | \(1/(2\sqrt{x})\) | \(x>0\) |
| \(\sin x\) | \(\cos x\) | \(\mathbb R\), radians |
| \(\cos x\) | \(-\sin x\) | \(\mathbb R\), radians |
| \(e^x\) | \(e^x\) | \(\mathbb R\) |
A demanding derivative without expanding
Differentiate
\[F(x)=\frac{e^{\sin(x^2)}(1+x^2)^3}{\sqrt{2+\cos x}}.\]First check the domain: \(2+\cos x\ge1\), so \(F\) is defined and positive for every real \(x\). Treat it as \(ABC\), where
\[A=e^{\sin(x^2)},\qquad B=(1+x^2)^3,\qquad C=(2+\cos x)^{-1/2}.\]Repeated use of the chain rule gives
\[A'=A\,2x\cos(x^2),\quad B'=6x(1+x^2)^2,\quad C'=\frac{\sin x}{2(2+\cos x)^{3/2}}.\]The rule for \(C\) follows by combining the reciprocal and square-root rules. The three-factor product rule gives \(F'=A'BC+AB'C+ABC'\). Factor \(ABC=F\):
\[F'(x)=F(x)\left[ 2x\cos(x^2)+\frac{6x}{1+x^2} +\frac{\sin x}{2(2+\cos x)} \right].\]Each term has a visible source. The factored answer is often more useful, and easier to check, than an expanded one.
Problems: combine and justify the rules
3.1 — Track every layer. Differentiate
\[H(x)=\frac{\sin(e^{x^2})}{1+\cos^2x}\]and explain why the formula is valid for every real \(x\). Identify the inner-function factors in your answer.
Hint
Differentiate the numerator and denominator separately before applying the quotient rule.
Solution
\[H'(x)=\frac{ 2xe^{x^2}\cos(e^{x^2})(1+\cos^2x) +2\sin(e^{x^2})\sin x\cos x }{(1+\cos^2x)^2}.\]The numerator’s nested chain contributes \(2x\) and \(e^{x^2}\). The denominator derivative is \(-2\cos x\sin x\), producing the positive second term. Since \(1+\cos^2x\ge1\), there are no excluded real points.
3.2 — A missing point in disguise. Let \(q(x)=(x^2-1)/(x-1)\) for \(x\ne1\). Compute \(q'\) on its domain. Then choose \(q(1)\) so the extended function is differentiable at \(1\), and find that derivative. Why can you not substitute \(x=1\) into the original quotient-rule expression?
Hint
Simplify the function before differentiating it.
Solution
For \(x\ne1\), \(q(x)=x+1\), so \(q'(x)=1\). Continuity requires \(q(1)=2\), and that extension is the linear function \(x+1\) everywhere, with derivative \(1\) at the join. The original function has no value at \(1\) and the original denominator vanishes there; only the extension supplies a derivative at that point.
3.3 — The converse of the product rule fails. Construct two functions, both nondifferentiable at zero, whose product is differentiable there. Next construct a function \(f\) that is nondifferentiable at zero while \(f^2\) is differentiable there. Explain why these examples do not contradict the product rule.
Hint
Try \(f(x)=\lvert x\rvert\).
Solution
Take \(f(x)=g(x)=\lvert x\rvert\). Both fail to be differentiable at zero, but \(fg=f^2=x^2\) has derivative zero there. The rule is a sufficient implication from differentiability of the factors; it does not claim that differentiability of a product forces differentiability of its factors.
3.4 — Repeated differentiation. Define \(f^{(0)}=f\) and \(f^{(n+1)}=(f^{(n)})'\) whenever these functions exist. For \(f(x)=xe^x\), calculate the first three derivatives, conjecture a formula for \(f^{(n)}\), and prove it by induction.
Hint
Keep each answer factored by \(e^x\).
Solution
The first three are \((x+1)e^x\), \((x+2)e^x\), and \((x+3)e^x\). The formula \(f^{(n)}(x)=(x+n)e^x\) holds for every integer \(n\ge0\). Differentiating the formula gives \(e^x+(x+n)e^x=(x+n+1)e^x\), proving the induction step.
Checkpoint. Given a complicated expression, describe its outermost operation before differentiating. State any restrictions on denominators and roots. A short answer is useful only if every factor has been accounted for.