MAT1001 Differential Calculus: Lecture Notes

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Chapter 11 Extra Proofs and Derivations

The material in this chapter is not examinable and is included for completeness. Here we will give a selection of proof and derivations for results and formulas that were used in the rest of the notes. You could consider this whole chapter to be one big mathematical deviation.

To prove the product rule from first principles we proceed as follows. Consider \(f(x)=p(x)q(x)\) then

\begin{align*} \frac {\ud f}{\ud x} &=\lim _{h\to 0}\frac {f(x+h)-f(x)}{h}\\ &=\lim _{h\to 0}\frac {p(x+h)q(x+h)-p(x)q(x)}{h}\\ &=\lim _{h\to 0}\frac {p(x+h)q(x+h)-p(x+h)q(x)+p(x+h)q(x)-p(x)q(x)}{h}\\ &=\lim _{h\to 0}\frac {p(x+h)\left (q(x+h)-g(x)\right )}{h}+\lim _{h\to 0}\frac {\left (p(x+h)-p(x)\right )q(x)}{h}\\ &=\lim _{h\to 0}p(x+h)\frac {q(x+h)-q(x)}{h}+q(x)\lim _{h\to 0}\frac {p(x+h)-p(x)}{h}\\ &=p(x)\lim _{h\to 0}\frac {q(x+h)-q(x)}{h}+q(x)\lim _{h\to 0}\frac {p(x+h)-p(x)}{h}\\ &=p(x)\frac {\ud q}{\ud x}+q(x)\frac {\ud p}{\ud x}, \end{align*} which is the product rule.

For the quotient rule, we can either prove it from first principles, which is fiddly, or we can use the product rule on \(f(x)=p(x)g(x)\) where \(g(x)=1/q(x)\). Here we will take the second approach as hopefully that will be easier to follow. Consider \(f(x)=p(x)/q(x)\) and let \(g(x)=1/q(x)\) then applying the product rule means that

\begin{align*} \frac {\ud f}{\ud x} &=\frac {\ud }{\ud x}\left (p(x)g(x)\right )\\ &=p(x)\frac {\ud g}{\ud x}+g(x)\frac {\ud p}{\ud x}\\ &=p(x)\frac {\ud }{\ud x}\left (q^{-1}\right )+\frac {1}{q(x)}\frac {\ud p}{\ud x}\\ &=-p(x)q^{-2}\frac {\ud q}{\ud x}+\frac {1}{q(x)}\frac {\ud p}{\ud x}\\ &=\frac {1}{q^{2}}\left (q(x)\frac {\ud p}{\ud x}-p(x)\frac {\ud q}{\ud x}\right ). \end{align*}

Mathematical Diversion 11.1. Note that in the above calculation we have used the chain rule, \(\left ((f\circ g)(x)\right )'=f'(g(x))g'(x)\), which we did not discuss until after we had introduced the quotient rule. Also, note that if we were mathematicians we would need to carefully think about when \(p(x)/q(x)\) is differentiable, and implementing the product rule does not need that, it just gives that \(p(x)(q(x))^{-1}\) is differentiable if \(p(x)\) and \((q(x))^{-1}\) is. This is why for mathematicians, they would prove the quotient rule using differentiation from first principles, or using logarithmic differentiation. The interested reader should have a look at the Proof of various derivative properties section of [Dawkins, 2025a] to see the proof in this way.

Next, following [Dawkins, 2025a], we prove the chain rule as follows: let \(y=f(u), u=g(x)\) then we know that

\begin{equation*} \frac {\ud u}{\ud x}=\lim _{h\to 0}\frac {u(x+h)-u(x)}{h}, \end{equation*}

and that

\begin{equation*} \lim _{h\to 0}\left (\frac {u(x+h)-u(x)}{h}-\frac {\ud u}{\ud x}\right )=\lim _{h\to 0}\frac {u(x+h)-u(x)}{h}-\lim _{h\to 0}\frac {\ud u}{\ud x}=\frac {\ud u}{\ud x}-\frac {\ud u}{\ud x}=0., \end{equation*}

Now we can define

Chapter 12 Standard Derivatives and Integrals

There are several functions that it is worth knowing the derivatives and integrals of. When we first introduced their derivatives we either derived them from first principles or used techniques like the product rule, chain rule, or integration by parts, to derive them from already known results. However, this takes time. You do not need to memorise the following list of standard derivatives and integrals, but you may find the list to be a useful resource when going through the tutorial problems or when revising for the exam.

Remember, you should still be able to derive these results if you need to, the list is just intended as an aid.

12.1 Derivatives

Table 12.1: Table of standard derivatives

.
\(y=f(x)\)	\(\frac {\ud y}{\ud x}=f^{\prime }(x) \)
\(n\) constant	\(0\)
\(x\)	\(1\)
\(x^{n}\), \(n\) constant	\(nx^{n-1}\)
\(e^{kx}\), \(k\) constant	\(ke^{kx}\)
\(\ln (x)\)	\(\frac {1}{x}\)
\(\sin (k x)\)	\(k\cos (k x)\)
\(\cos (kx)\)	\(-k\sin (kx)\)
\(\tan (kx)\)	\(k\sec ^{2}(kx)\)
\(\arcsin (x)\)	\(\frac {1}{\sqrt {1-x^{2}}} \)
\(\arccos (x)\)	\(-\frac {1}{\sqrt {1-x^{2}}}\)
\(\sinh (x)\)	\(\cosh (x)\)
\(\cosh (x)\)	\(\sinh (x)\)
\(\tanh (x)\)	\(\sech ^{2}(x)\)
\(f(x)g(x)\)	\(f^{\prime }(x)g(x)+f(x)f^{\prime }(x)\)
\(\frac {f(x)}{g(x)}\)	\(\frac {f^{\prime }(x)g(x)-f(x)f^{\prime }(x)}{(g(x))^{2}} \)

12.2 Integrals

Table 12.2: Table of standard integrals

.
\(y=f(x)\)	\(\int f(x) \ud x\)
constant \(k\)	\(kx+c\)
\(x^{n}\)	\(\frac {x^{n+1}}{n+1}+c\)
\(\frac {1}{x}\)	\(\ln (\vert x\vert )+c\)
\(e^{kx}\), \(k\) constant	\(\frac {e^{kx}}{k} +c\)
\(x^{-n}\)	\(\frac {x^{-n+1}}{-n+1} +c\), \(n\neq 1\)
\(\cos (x)\)	\(\sin (x) +c\)
\(\sin (x)\)	\(-\cos (x)+c\)
\(f(x)g^{\prime }(x)\)	\(f(x)g(x)-\int f^{\prime }(x)g(x)\ud x \)