Proposition 1.6.
Suppose \(f(x)\) and \(g(x)\) are differentiable at \(x=x_0\) then so are their sum and product. Moreover,
\begin{align*}
(f+g)'(x_0) & = f'(x_0) + g'(x_0)\\
(fg)'(x_0) & = f'(x_0)g(x_0) + f(x_0)g'(x_0)
\end{align*}
Section 1.2 Derivatives
In this section, we study various operations on differentiable functions that preserve differentiability. The derivative of a function obtained by these operations can be expressed in terms of its constituent functions and their derivatives. These expressions are the so-called "rules of differentiation".
Subsection 1.2.1 Rules of Differentiation
Viewing these derivatives as functions, we write
\begin{align*}
(f+g)' & = f' + g' \tag{sum rule} \\
(fg)' & = f'g+ fg' \tag{product rule}
\end{align*}
Since the derivative of a constant \(c\) is zero, in particular
\begin{align*}
(f+c)' & = f' \\
(cf)' & = cf'+ c'f =cf'
\end{align*}
Also since \(f-g = f+(-1)g\text{,}\) \((f-g)' = f' + ((-1)g)' = f'-g'\text{.}\)
Example 1.7.
Applying the product rule to \(f(x)=g(x)=x\text{,}\) we conclude that \((fg)(x) = x^2\) is differentiable. Moreover,
\begin{equation*}
(x^2)' = x'(x) + x(x') = 2x.
\end{equation*}
That is exactly what we have found using the definition.
More generally, one can check by induction that
\begin{gather*}
(x^n)' = nx^{n-1}
\end{gather*}
for every positive integer \(n\text{.}\)
In fact, not only the non-negative powers of \(x\) but for any real number \(r\text{,}\) the function \(x \mapsto x^r\) is differentiable and that
\begin{gather*}
(x^r)' = rx^{r-1} \tag{power rule}
\end{gather*}
We will be justifying this later.
Since polynomial functions can be obtained from constants and the identity function by taking sum and product, they are differentiable. Moreover,
\begin{equation}
\left( \sum_{k=0}^n a_kx^k\right)' = \sum_{k=0}^n ka_{k}x^{k-1}\tag{1.3}
\end{equation}
Example 1.8.
Let us verify just one instant of the power rule. For \(x, x_0 \gt
0\text{,}\) as \(x-x_0 = (\sqrt{x}+\sqrt{x_0})(\sqrt{x}-\sqrt{x_0})\text{,}\)
\begin{equation*}
\sqrt{x}-\sqrt{x_0} = \frac{1}{\sqrt{x}+\sqrt{x_0}}(x-x_0)
\end{equation*}
The function \(\varphi_{x_0}(x) = \dfrac{1}{\sqrt{x}+\sqrt{x_0}}\) is continuous at \(x_0\text{.}\) Thus, the derivative of \(1/x\) at \(x_0\) is \(-\dfrac{1}{2\sqrt{x_0}}\) and this verifies the power rule in the case \(r=1/2\text{.}\)
A fundamental operation on functions is composition. One uses the Chain Rule to compute the derivative of a composition.
Proposition 1.9. The Chain Rule.
Suppose \(f(x)\) is differentiable at \(x=x_0\) and \(g(u)\) is differentiable at \(f(x_0)\) then the composition \(g \circ
f(x)\) is differentiable at \(x_0\text{.}\) Moreover,
\begin{equation*}
(g \circ f)'(x_0) = g'(f(x_0))f'(x_0).
\end{equation*}
Proof.
It follows from the assumptions that there are functions \(\gamma(u)\) and \(\varphi(x)\) continuous at \(u_0 = f(x_0)\) and \(x_0\text{,}\) respective such that
\begin{gather}
f(x) - f(x_0) = \varphi(x)(x-x_0)\tag{1.4}\\
g(u) - g(u_0) = \gamma(u)(u-u_0)\tag{1.5}
\end{gather}
\begin{align*}
g(f(x)) - g(f(x_0)) & = \gamma(f(x))(f(x)-f(x_0))\\
& = \gamma(f(x))\varphi(x)(x-x_0)
\end{align*}
Since \(f(x)\) is differentiable at \(x_0\text{,}\) it is continuous at \(x_0\) as well Proposition 1.5. So, we conclude that the function \(x \mapsto \gamma(f(x))\varphi(x)\) is continuous at \(x_0\) since continuous function are closed under composition and product. This shows that \(g\circ f\) is differentiable at \(x_0\text{,}\) moreover
\begin{equation*}
(g\circ f)'(x_0) = \gamma(f(x_0))\varphi(x_0) = g'(f(x_0))f'(x_0).
\end{equation*}
In Leibniz’s notation, the Chain Rule is expressed as
\begin{equation}
\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}.\tag{1.6}
\end{equation}
The two notations are related as follows. Let \(y\) be the variable depending on \(x\) in the way specified by \(g \circ f\text{,}\) i.e. let \(y = g(f(x))\text{.}\) We may not recognize the derivative of \(y\) with respect to \(x\) immediately. However, if we think of \(f(x)\) as a variable \(u\) and \(g\text{,}\) as a function of \(u\text{,}\) is differentiable, then \(dy/dx = (g\circ f)'(x)\text{,}\) \(du/dx = f'(x)\) and \(dy/du = g'(u)\) and the Chain Rule in Leibniz notation changes back to
\begin{equation*}
(g \circ f)'(x) = g'(u)f'(x) = g'(f(x))f'(x).
\end{equation*}
From this point of view, the Chain Rule is a way of finding derivatives by substitution. And if we recognize \(dy/du = g'(u)\text{,}\) that is the derivative of the "last step" \(g\) with respect to \(u\text{,}\) then we reduce the problem of finding the derivative of the composition \(g
\circ f(x)\) to an easier problem of finding the derivative \(du/dx =
f'(x)\text{.}\) Because of this, the Chain Rule is so called the "onion rule" (peeling an onion from the outer-most layer first then work ones way to the inner part, one layer at a time). Thus, in applying the Chain Rule, it is crucial to understand how the composition is built step-by-step.
Example 1.10.
Let us compute the derivative of \(y=\frac{1}{1-\sqrt{x^2+1}}\) by the Chain Rule (from the substitution point of view). First, the last step of this composition is taking the reciprocal. Thus, we let \(y=1/u\) and \(u=1-\sqrt{x^2+1}\text{,}\) so
\begin{equation*}
\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} =
-\frac{1}{u^2}\frac{du}{dx}.
\end{equation*}
Thus, our first application of the Chain Rule reduces the original problem to the one of finding the derivative of a simpler function \(u=1-\sqrt{x^2+1}\text{.}\) We apply the Chain Rule again, this time we recognize the derivative of the square root function. And so we make the substitution \(v=x^2+1\text{.}\) And so \(u=1-\sqrt{v}\text{.}\) According to the Chain Rule,
\begin{equation*}
\frac{du}{dx} = \frac{du}{dv}\frac{dv}{dx}
= \frac{1-\sqrt{v}}{dv}\frac{dv}{dx} =
-\frac{1}{2\sqrt{v}}\frac{dv}{dx}.
\end{equation*}
Finally, we recognize the derivative of \(v=x^2+1\) with respect to \(x\) as \(dv/dx = d(x^2+1)/dx =2x\text{.}\) Putting these back together, we have
\begin{equation*}
\frac{dy}{dx}
=\left(-\frac{1}{u^2}\right)\left(-\frac{1}{2\sqrt{v}}\right)(2x) =
\frac{x}{\sqrt{x^2+1}(1-\sqrt{x^2+1})^2}.
\end{equation*}
Subsection 1.2.2 Implicit Differentiation
The curve \(y^2 = x\) is not the graph of a function of \(x\) because, for instance, the points \((1,-1)\) and \((1,1)\) on the graph have different \(y\) coordinates but the same \(x\) coordinate.
However, locally around any fixed point \((x_0,y_0)\) with \(x_0 \neq 0\) the curve is indeed the graph of some function of \(x\text{.}\) This is the simplest case of the Implicit Function Theorem. For example, around \((1,1)\) the curve is the graph of \(y = \sqrt{x}\) and around \((1,-1)\) it is the graph of \(y
= -\sqrt{x}\text{.}\)
Suppose the equation \(F(x,y) = 0\) defines \(y\) as a function of \(x\) locally near a point \((x_0,y_0)\text{.}\) Assuming \(y(x)\) is differentiable at \(x=x_0\text{,}\) we can try to find the derivative of \(y(x)\) at \(x=x_0\) without explicitly writing \(y(x)\) as an expression of \(x\) which is often cumbersome, if not impossible to do. Let us illustrate this by our parabola example \(y^2 = x \text{.}\)
Example 1.12.
Let us compute the slope of the tangent to the curve \(y^2=x\) (or \(y^2-x = 0\)) at the point \((x_0,y_0) = (1,-1)\text{.}\) We do this not by solving \(y\) explicitly as \(y(x)=-\sqrt{x}\) then find the derivative at \(x=1\) but by differentiating the relation \(y^2 =x\text{.}\) So we have
\begin{align*}
\frac{d}{dx} y^2 & =\frac{d}{dx}x\\
\frac{d}{dy}y^2 \dfrac{dy}{dx} & = 1\\
2y\dfrac{dy}{dx} & = 1
\end{align*}
Since \((1,-1)\) is on the graph of \(y(x)\text{,}\) \(y(1)
= -1\) and so \(1= 2y(1)y'(1) = 2(-1)y'(1)\text{.}\) That is \(y'(1) = -1/2\text{.}\) Also, from the relation \(2y(x)y'(x)
=1\) we must conclude \(y'(x) = 1/(2y(x))\text{.}\) Therefore, if we know \(y\) explicitly as \(-\sqrt{x}\text{,}\) then we can also express \(y'(x)\) explicitly as \(-1/(2\sqrt{x})\text{.}\)
