17.5: Appendix E- Proofs

This Appendix was written by Dr. Sophie Burrill.

E.1 Proof of the power rule

We present a proof for the power rule. Recall from Section 4.1:

The power rule states that the derivative of the power function \(f(x)=x^{n}\) is \(n x^{n-1}\).

Proof. We begin with the definition of the derivative:

\[f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{(x+h)^{n}-x^{n}}{h} . \nonumber \]

As mentioned in Section 4.1, the binomial \((x+h)^{n}\) entails lenthy algebra. We employ the binomial theorem (which we do not prove):

Binomial theorem. If \(n\) is a positive integer, then

\[\begin{aligned} (x+y)^{n}= & x^{n}+n x^{n-1} y+\frac{n(n-1)}{2 \cdot 1} x^{n-2} y^{2}+\frac{n(n-1)(n-2)}{3 \cdot 2 \cdot 1} x^{n-3} y^{3} \\ & +\ldots+\frac{n(n-1) \ldots(n-k+1)}{k(k-1) \ldots 2 \cdot 1} x^{n-k} y^{k}+\ldots+n x y^{n-1}+y^{n} \end{aligned} \nonumber \]

Note that this means the expansion of \((x+y)^{n}\) is a sum of terms of the form

\[c_{k} x^{n-k} y^{k}, \quad k=0,1, \ldots, n \nonumber \]

where \(c_{k}\) (called a "binomial coefficient") is a coefficient that depends on both \(n\) and \(k\). The exact form of \(c_{k}\) is not necessary for the proof of the power rule - except for the terms \(c_{0}=1\) and \(c_{1}=n\), the coefficients of \(x^{n}\) and \(x^{n-1} y\).

Let us use the binomial theorem and expand the numerator in the defini- tion of the derivative:

\[\begin{aligned} f(x+h)-f(x)= & (x+h)^{n}-x^{n} \\ = & \left(c_{0} x^{n}+c_{1} x^{n-1} h+c_{2} x^{n-2} h^{2}+c_{3} x^{n-3} h^{3}\right. \\ & \left.+\ldots+c_{n-1} x h^{n-1}+c_{n} h^{n}\right)-x^{n} \end{aligned} \nonumber \]

We can rewrite using the fact that \(c_{0}=1\) and note that the terms \(x^{n}\) cancel:

\[(x+h)^{n}-x^{n}=\left(x^{n}+c_{1} x^{n-1} h+\ldots+c_{n} h^{n}\right)-x^{n}=c_{1} x^{n-1} h+\ldots+c_{n} h^{n} \nonumber \]

All of the remaining terms have \(h\) as a factor, which we can factor out:

\[(x+h)^{n}-x^{n}=h\left[c_{1} x^{n-1}+c_{2} x^{n-2} h+\cdots+c_{n} h^{n-1}\right] \nonumber \]

Substituting this into the definition of the derivative we achieve:

\[\begin{aligned} f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{(x+h)^{n}-x^{n}}{h} \\ & =\lim _{h \rightarrow 0} \frac{h\left[c_{1} x^{n-1}+c_{2} x^{n-2} h+\cdots+c_{n} h^{n-1}\right]}{h} \\ & =\lim _{h \rightarrow 0}\left[c_{1} x^{n-1}+c_{2} x^{n-2} h+\cdots+c_{n} h^{n-1}\right] \end{aligned} \nonumber \]

Note that all terms except for the first have \(h\) as a factor and so tend to 0 as \(h \rightarrow 0\). This gives that

\[f^{\prime}(x)=c_{1} x^{n-1}, \nonumber \]

and we have already noted that \(c_{1}=n\), so

\[f^{\prime}(x)=n x^{n-1}, \nonumber \]

which proves the power rule.

E.2 Proof of the product rule

We proof the product rule. Recall from Section \(4.1\) :

The product rule: If \(f(x)\) and \(g(x)\) are two functions, each differentiable in the domain of interest, then

\[\frac{d[f(x) g(x)]}{d x}=\frac{d f(x)}{d x} g(x)+\frac{d g(x)}{d x} f(x) \nonumber \]

Another notation for this rule is

\[[f(x) g(x)]^{\prime}=f^{\prime}(x) g(x)+g^{\prime}(x) f(x) \nonumber \]

Proof. Let \(k(x)=f(x) g(x)\), the product of the two functions. We use the definition of the derivative:

\[k^{\prime}(x)=\lim _{h \rightarrow 0} \frac{k(x+h)-k(x)}{h}=\lim _{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x) g(x)}{h} \nonumber \]

Adding \(0=f(x) g(x+h)-f(x) g(x+h)\) allows us to perform some helpful factoring:

\[\begin{aligned} k^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x) g(x+h)+f(x) g(x+h)-f(x) g(x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{[f(x+h)-f(x)] \cdot g(x+h)+f(x) \cdot[g(x+h)-g(x)]}{h} \\ & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} g(x+h)+f(x) \frac{g(x+h)-g(x)}{h} \end{aligned} \nonumber \]

Due to properties of limits (see Appendix D.2) we can distribute the limit and recognize familiar derivatives:

\[\begin{aligned} k^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \lim _{h \rightarrow 0} g(x+h)+\lim _{h \rightarrow 0} f(x) \lim _{h \rightarrow 0} \frac{g(x+h)-g(x)}{h} \\ & =f^{\prime}(x) g(x)+f(x) g^{\prime}(x) . \end{aligned} \nonumber \]

Thus we have proved the power rule, that

\[[f(x) g(x)]^{\prime}=f^{\prime}(x) g(x)+f(x) g^{\prime}(x) \nonumber \]

E.3 Proof of the quotient rule

We provide a proof the quotient rule. Recall from Section 4.1:

The quotient rule: If \(f(x)\) and \(g(x)\) are two functions, each differentiable in the domain of interest, then

\[\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\frac{\frac{d f(x)}{d x} g(x)-\frac{d g(x)}{d x} f(x)}{[g(x)]^{2}} \nonumber \]

We can also write this in the form

\[\left[\frac{f(x)}{g(x)}\right]^{\prime}=\frac{f^{\prime}(x) g(x)-g^{\prime}(x) f(x)}{[g(x)]^{2}} \nonumber \]

Proof. This proof also follows from the definition of the derivative; it contains some careful arithmetic. Let \(k(x)=\frac{f(x)}{g(x)}\). Using the definition of the derivative we get:

\(k^{\prime}(x)=\lim _{h \rightarrow 0} \frac{k(x+h)-k(x)}{h}=\lim _{h \rightarrow 0} \frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}\right]\). Finding a common denominator and then adding \(0=g(x+h) f(x+h)-g(x+\) h) \(f(x+h)\) in the numerator we proceed:

\[\begin{aligned} k^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{f(x+h) g(x)-f(x) g(x+h)}{g(x+h) g(x)}\right] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{f(x+h) g(x)-f(x+h) g(x+h)+f(x+h) g(x+h)-f(x) g(x+h)}{g(x+h) g(x)}\right] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-f(x+h)[g(x+h)-g(x)]+g(x+h)[f(x+h)-f(x)]}{g(x+h) g(x)}\right] \end{aligned} \nonumber \]

Using properties of limits and identifying the definition of the derivative for \(g^{\prime}(x)\) and \(f^{\prime}(x)\) leads us to:

\[\begin{aligned} k^{\prime}(x) & =\lim _{h \rightarrow 0}\left[\frac{-f(x+h) g^{\prime}(x)}{g(x+h) g(x)}+\frac{g(x+h) f^{\prime}(x)}{g(x+h) g(x)}\right] \\ & =\frac{-f(x) g^{\prime}(x)+g(x) f^{\prime}(x)}{[g(x)]^{2}}=\frac{f^{\prime}(x) g(x)-g^{\prime}(x) f(x)}{[g(x)]^{2}} . \end{aligned} \nonumber \]

We have thus proved the quotient rule. Despite the arithmetic required, hopefully the fact that the definition of the derivative is all that is required provides the reader with some comfort.

E.4 Proof of the chain rule

We present a plausibility argument for the chain rule. Recall from Section 8.1:

If \(y=g(u)\) and \(u=f(x)\) are both differentiable functions and \(y=g(f(x))\) is the composite function, then the chain rule of differentiation states that

\[\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d x} . \nonumber \]

Proof. We first note that if a function is differentiable, it is also continuous. Because of this continuity, when \(x\) changes a very little, \(u\) can change only by a little - there are no abrupt jumps. Thus, using our notation, if \(\Delta x \rightarrow 0\) then \(\Delta u \rightarrow 0\).

Now consider the definition of the derivative \(d y / d u\) :

\[\frac{d y}{d u}=\lim _{\Delta u \rightarrow 0} \frac{\Delta y}{\Delta u} . \nonumber \]

This means that for any (finite) \(\Delta u\),

\[\frac{\Delta y}{\Delta u}=\frac{d y}{d u}+\varepsilon, \nonumber \]

where \(\varepsilon \rightarrow 0\) as \(\Delta u \rightarrow 0\). Then

\[\Delta y=\frac{d y}{d u} \Delta u+\varepsilon \Delta u . \nonumber \]

Now divide both sides by some (nonzero) \(\Delta x\) :

\[\frac{\Delta y}{\Delta x}=\frac{d y}{d u} \frac{\Delta u}{\Delta x}+\varepsilon \frac{\Delta u}{\Delta x} \nonumber \]

Taking \(\Delta x \rightarrow 0\) we get \(\Delta u \rightarrow 0\), (by continuity) and hence also \(\varepsilon \rightarrow 0\) so that as desired,

\[\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d x} \nonumber \]