## Tag Archives: derivatives

### Differentiation — continued

Differentiation of the Sum, Difference, Product and Quotient of Two Functions

Theorem:

If $f, g: I \rightarrow \Re$ are differentiable at $x_{0} \in I$, where I is an open interval in $\Re$, then so are $f \pm g$, fg at $x_{0}$. Furthermore, $\frac{f}{g}$ is differentiable at $x_{0}$ if $g(x_{0}) \neq 0$. We have

(a) derivative of $f \pm g$ at $x_{0}$ is $f^{'}(x_{0}) \pm g^{'}(x_{0})$

(b) derivative of $f.g$ at $x_{0}$ is $f^{'}(x_{0})g(x_{0})+f(x_{0})g^{'}(x_{0})$

(c) derivative of $\frac{f}{g}$ at $x_{0}$ is $\frac{f^{'}(x_{0})g(x_{0})-f(x_{0})g^{'}(x_{0})}{{g(x_{0})}^{2}}$ if $g^{'}(x_{0}) \neq 0$

The proofs are straightforward and therefore omitted.

We also have

Theorem (Chain Rule):

Let I and J be two intervals in $\Re$ and $f: I \rightarrow J$, and $g: J \rightarrow \Re$ be differentiable at $x_{0} \in I$ and $f(x_{0}) \in J$ respectively. Then, $h \equiv g \circ f : I \rightarrow \Re$ is differentiable at $x_{0}$ and

$h^{'}(x_{0})=g^{'}(f(x_{0}))f^{'}(x_{0})$

Note that $h = g \circ f$ is defined as $h(x)=g(f(x))$.

Proof.

Let us write $y=f(x)$ so that by the continuity of f at $x_{0}$, we have that as $x \rightarrow x_{0}$, $y \rightarrow y_{0}=f(x_{0})$. Since g is differentiable at $y_{0}$, we have

$g(y)-g(y_{0})=(g^{'}(y_{0})+r_{1}(y,y_{0}))(y-y_{0})$

Here $r_{1}(y,y_{0})$ as $y \rightarrow y_{0}$. Again, since f is differentiable at $x_{0}$, we have

$f(x)-f(x_{0})=(f^{'}(x_{0})+r_{2}(x,x_{0}))(x-x_{0})$.

Here, $r_{2}(x,x_{0}) \rightarrow 0$ as $x \rightarrow x_{0}$. Thus, we have

$f(x)-g(f(x_{0})) = (g^{'}(f(x_{0}))+r_{1}(y,y_{0}))(f(x)-f(x_{0}))$, which equals

$(g^{'}(y_{0})+r_{1}(y,y_{0}))(f^{'}(x_{0})+r_{2}(x,x_{0}))(x-x_{0})$, which in turn, equals

$g^{'}(f(x_{0}))f^{'}(x_{0})(x-x_{0})+(x-x_{0})r_{3}$ where

$r_{3}=g^{'}(f(x_{0}))r_{2}(x,x_{0})+f^{x_{0}}r_{1}(y,y_{0})+r_{1}(y,y_{0})r_{2}(x,x_{0})$.

Surely, $r_{3} \rightarrow 0$ as $x \rightarrow x_{0}$ and hence,

$h^{'}(x_{0})=g^{'}(f(x_{0}))f^{'}(x_{0})$.

The above result is also often called the Chain Rule.

Differential Notation of Leibniz

For a differentiable function f, if we write $y=f(x)$ and $y+\triangle y=f(x+\triangle x)$, then we get

$\lim_{\triangle x \rightarrow 0} \frac{\triangle y}{\triangle x}=\lim_{\triangle x \rightarrow 0}\frac{f(x+\triangle x)-f(x)}{\triangle x}=f^{'}(x)$

The expression

$\lim_{\triangle x \rightarrow 0}\frac{\triangle y}{\triangle x}$ is often written as $\frac{dy}{dx}$.

It is NOT true that $\frac{dy}{dx}$ is the quotient of limits of $\triangle y$ and of $\triangle x$ because both of them tend to zero. It should rather be thought of as an operator (or operation) $\frac{d}{dx}$ is the operation of differentiation operating on the variable y so that we have

$\frac{d}{dx}(y)=\frac{dy}{dx}$.

The operator $\frac{d}{dx}$ has the property $\frac{d}{dx}(y+u)=\frac{dy}{dx}+\frac{du}{dx}$ and

$\frac{d}{dx}(cy)=c\frac{dy}{dx}$, and for f and g, two differentiable functions with the domain of g containing the range of f, if we write $y=f(x)$ and $u=g(y)$, we have $u=g(f(x))$, then the chain rule gives $\frac{dy}{dx}=\frac{du}{dy}. \frac{dy}{dx}$. In the case, when $y=f(x)$ so that $\frac{dy}{dx}=f^{x}$ we write, following the German mathematician Leibnitz,

$dy=f^{'}(x)dx$,

and dy and dx are called the differentials of y and x respectively.

Remark:

Let $f: \Re^{2} \rightarrow \Re$ be a function. We say that f is differentiable at $(x_{0},y_{0}) \in \Re^{2}$ if we can find numbers A, B depending on $(x_{0},y_{0})$ only, so that

$f(x,y) - f(x_{0},y_{0})=A(x-x_{0})+B(y-y_{0})+R(x,y,x_{0},y_{0})$ such that

$\lim \frac{R(x,y,x_{0},y_{0})}{\sqrt{(x-x_{0})^{2}-(y-y_{0})^{2}}}=0$ as $(x,y) \rightarrow (x_{0},y_{0})$

Observe that

$A=\lim_{x \rightarrow x_{0}}\frac{f(x,y_{0})-f(x_{0},y_{0})}{x-x_{0}}$ and $B=\lim_{y \rightarrow y_{0}}\frac{f(x_{0},y)-f(x_{0},y_{0})}{y-y_{0}}$

We call A and B the partial derivatives of f with respect to x and y respectively at $(x_{0},y_{0})$, and we write

$A=\frac{\partial f}{\partial x}(x_{0},y_{0})$, $B=\frac{\partial f}{\partial y}(x_{0},y_{0})$

Sometimes, we also  write $\frac{\partial f}{\partial x}(x_{0},y_{0}) = f_{x}(x_{0},y_{0})$ and $\frac{\partial f}{\partial y}(x_{0},y_{0})=f_{y}(x_{0},y_{0})$. Again, as before, $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ may be thought of as operators which, operating on a function, give its partial derivatives with respect to x and y respectively.

Suppose that $\phi: \Re^{2} \rightarrow \Re$ is differentiable, and that x and $y: \Re \rightarrow \Re$ are differentiable functions. Furthermore, let $u: \Re \rightarrow \Re$ be defined by

$u(t)=\phi (x(t),y(t))$

It is not difficult to show that (Exercise!)

$u^{'}(t)=\frac{\partial \phi}{\partial x} x^{'}(t)+\frac{\partial \phi}{\partial y} y^{'}(t)$.

In Leibnitz’s notation, it reads $du=\frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y} dy$.

As an application of this idea, consider the notion of equipotential surface in electrostatics. An electrically charged (infinite) cylindrical conductor is one such and because of the symmetry, it is enough to look at only a horizontal section of the cylinder which is a closed curve, that is, the problem gets reduced to a 2-dimensional one. This equipotential curve is given by the equation $\phi (x,y)=constant$, say c, where $\phi$ is the real valued potential function. What is the electric field outside the curve? Using Leibnitz’s notation as described above, we can write $d\phi=\frac{\partial \phi}{\partial x}dx + \frac{\partial \phi}{\partial y}dy$.

Since $\phi$ is a constant, the relation above reads as

$\frac{\partial \phi}{\partial x} x^{'}(t) + \frac{\partial \phi}{\partial y} y^{'}(t)=0$,

that is, the vector $(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y})$ is orthogonal to the tangent vector $(x^{'}(t),y^{'}(t))$.

Recalling that the electric field E at a point $(x,y)$ outside has components $E_{1}=-\frac{\partial \phi}{\partial x}$ and $E_{2}=-\frac{\partial \phi}{\partial y}$, we conclude that the electric field E is along the outward normal. For example, for a right circular cylinder, the electric field is along the radial direction of every section.

We have seen that if f is differentiable at $x_{0}$, then $f(x)-f(x_{0})-f^{'}(x_{0})(x-x_{0})$ is of smaller order than $(x-x_{0})$ as $x \rightarrow x_{0}$.

Writing $\triangle y=f(x)-f(x_{0})$, $\triangle x = (x-x_{0})$ we get

$\triangle y = f^{'}(x_{0}) \triangle x$

which is of smaller order than $\triangle x$. In other words, we are claiming that the increment in y is proportional to the increment in x when it is *small*, which is the principle of proportional parts. We have put the word small within asterisks as it is a relative term. Let us consider an example.

Let $f(x) = \sqrt {x}$ with $x \geq 0$. Then, $f^{'}(x)=\frac{1}{2\sqrt{x}}$ for $x > 0$. So, $f(16)=4$, $f(17)=f(16)+f^{'}(16).1+r.1$, since $\triangle x = 1$. This gives us $\sqrt{17}=4+\frac {1}{8} + r = 4.125+r$, whereas computation on a hand calculator will give $\sqrt{17}=4.1231 \ldots$ This shows that $|r|<0.002$ and thus the differential approximates the increment in $f(x)$ correct at least to the second place of decimal.

More later,

Nalin Pithwa

### Differentiation

We have seen how the concept of continuity is naturally associated with attempts to model gradual changes. For example, consider the function $f: \Re \rightarrow \Re$ given by $f(x)=ax+b$, where change in $f(x)$ is proportional to the change in x. This simple looking function is often used to model many practical problems. One such case is given below:

Suppose 30 men working for 7 hours a day can complete a piece of work in 16 days. In how many days can 28 men working for 6 hours a day complete the work? It must be evident to most of the readers that the answer is $\frac{16 \times 7 \times 30}{28 \times 6}=20$ days.

(While solving this we have tacitly assumed that the amount of work done is proportional to the number of men working, to the number of hours each man works per day, and also to the number of days each man works. Similarly, Boyle’s law for ideal gases states that pressure remaining constant, the increase in volume of a mass of gas is proportional to the increase in temperature of the gas).

But, there are exceptions to this as well. Galileo discovered that the distance covered by a body, falling from rest, is proportional to the square of the time for which it has fallen, and the velocity is proportional to the square root of the distance through which it has fallen. Similarly, Kepler’s law tells us that the square of the period of the planet going round the sun is proportional to the cube of the mean distance from the sun.

These and many other problems involve functions that are not linear. If for example we plot the graph of the distance covered by a particle versus time, it is a straight line only when the motion is uniform. But, we are seldom lucky to encounter only uniform motion. (Besides, uniform motion would be so monotonous. Perhaps, there would be no life at all motions if all motions were uniform. Imagine a situation in which each body is in uniform motion. A body at rest would be eternally at rest and those once in motion, would never stop.) So the simple method of proportionality becomes quite inadequate to tackle such non-linear problems. The genius of Newton lay in looking at those problems which are next best to linear, the ones that are nearly linear.

We know that the graph of a linear function is a straight line. What Newton suggested was to look at functions, small portions of whose graphs look almost like a straight line (see Fig 1).

In Fig 1, the graph certainly is not a straight line. But a small portion of it looks like a straight like a straight line. To formalize this idea, we need the concept of differentiability.

Definition.

Let I be an open interval and $f: I \rightarrow \Re$ be a function. We say that f is locally linear or differentiable at $x_{0} \in I$ if there is a constant m such that

$f(x)-f(x_{0})=m(x-x_{0})+r(x_{0},x)(x-x_{0})$

or equivalently, for x in a punctured interval around $x_{0}$,

$\frac{f(x)-f(x_{0})}{x-x_{0}}=m+r(x_{0},x)$

where $r(x_{0},x) \rightarrow 0$ as $x \rightarrow x_{0}$

What this means is that for small enough $x-x_{0}$, $\frac{f(x)-f(x_{0})}{x-x_{0}}$ is nearly a constant or, equivalently, $f(x)-f(x_{0})$ is nearly proportional to the increment $x-x_{0}$. This is what is called the principal of proportional parts and used very often in calculations using tables, when the number for which we are looking up the table is not found there.

Thus, if a function f is differentiable at $x_{0}$, then $\lim_{x \rightarrow x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}$

exists and is called the derivative of f at $x_{0}$ and denoted by $f^{'}(x_{0})$. So we write

$\lim_{x \rightarrow x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}=f^{'}(x_{0})$.

We need to look at functions which are not differentiable at some point, to fix our ideas. For example, consider the function $f: \Re \rightarrow \Re$ defined by $f(x)=|x|$.

This function though continuous at every point is not differentiable at $latex x=0$. In fact, $\lim_{x \rightarrow 0_{+}}\frac{|x|}{x}=-1$. What all this means is that if one looks at the graph of $f(x)=|x|$, it has a sharp corner at the origin.

No matter how small a part of the graph containing the point $(0,0)$ is taken, it never looks like a line segment. The reader can test for the non-differentiability of $f(x)=|\sin{x}|$ at $x=n\pi$.

This leads us to the notion of the direction of the graph at a point: Suppose $f: I \rightarrow \Re$ is a function differentiable at $x_{0} \rightarrow I$, and let P and Q be the points $(x_{0},f(x_{0}))$ and $(x, f(x))$ respectively in the graph of f. (see Fig 2).

The chord PQ has the slope $\frac{f(x)-f(x_{0})}{x-x_{0}}$. As x comes close to $x_{0}$, the chord tends to the tangent to the curve at $(x_{0}, f(x_{0}))$. So, $\lim_{x \rightarrow x_{0}} \frac{f(x)-f(x_{0}}{x-x_{0}}$ really represents the slope of the tangent at $(x_{0},f(x_{0}))$ (see Fig 3).

Similarly, if $x(t)$ is the position of a moving point in a straight line at time t, then $\frac{x(t)-x(t_{0}}{t-t_{0}}$ is its average velocity in the interval of time $[t_{0},t]$. Its limit as t goes to $t_{0}$, if it exists, will be its instantaneous velocity at the instant of time $t_{0}$. We have

$x^{'}{t_{0}}=\lim_{t \rightarrow t_{0}}\frac{x(t)-x(t_{0})}{t-t_{0}}$ is instantaneous velocity at $t_{0}$.

If the limit of $\frac{f(x)-f(x_{0})}{x-x_{0}}$ does not exist as x tends to $x_{0}$, the curve $(x, f(x))$ cannot have a tangent at $(x_{0},f(x_{0}))$, as we saw in the case of $f(x)=|x|$ at $(0,0)$; the graph abruptly changes its direction. If we look at the motion of a particle which is moving with uniform velocity till time $t_{0}$ and is abruptly brought to rest at that instant, then its graph would look as in Fig 4a.

This is also what we think happens when a perfectly elastic ball impinges on another ball of the same mass at rest, or  when a perfectly elastic ball moving at a constant speed impinges on a hard surface (see fig 4b). We see that there is a sharp turn in the space time graph of such a motion at time $t=t_{0}$. Recalling the interpretation of

$x^{'}(t)=\lim_{t \rightarrow t_{0}} \frac{x(t)-x(t_{0})}{t-t_{0}}$ as its instantaneous velocity at $t=t_{0}$, we see that in the situation described above, instantaneous velocity at $t=t_{0}$ is not a meaningful concept.

We have already seen that continuous functions need not be differentiable at some points of their domain. Actually there are continuous functions which are not differentiable anywhere also. On the other hand, as the following result shows, every differentiable function is always continuous.

Theorem:

If a function is differentiable at $x_{0}$, then it is continuous there.

Proof:

If f is differentiable at $x_{0}$, then let $\lim_{x \rightarrow x_{0}} \frac{f(x)-f(x_{0}}{x-x_{0}}=l$. Setting

$r(x,x_{0})=\frac{f(x)-f(x_{0})}{x-x_{0}}-l$, we see that $\lim_{x \rightarrow x_{0}}r(x, x_{0})=0$. Thus, we have

$f(x)-f(x_{0})=(x-x_{0})l + (x-x_{0})r(x,x_{0})$

Now, $\lim_{x \rightarrow x_{0}} (f(x)-f(x_{0}))=\lim_{x \rightarrow x_{0}}(x-x_{0})l + \lim_{x \rightarrow x_{0}} (x-x_{0})r(x, x_{0})=0$

This shows that f is continuous at $x_{0}$.

QED.

Continuity of f at $x_{0}$ tells us that $f(x)-f(x_{0})$ tends to zero as $x - x_{0}$ tends to zero. But, in the case of differentiability, $f(x)-f(x_{0})$ tends to zero at least as fast as $x-x_{0}$. The portion $l(x-x_{0})$ goes to zero no doubt but the remainder $|f(x)-f(x_{0})-l(x-x_{0})|$ goes to zero at a rate faster than that of $|x-x_{0}|$. This is how differentiation was conceived by Newton and Leibniz. They introduced a concept called an infinitesimal. Their idea was that when $x-x_{0}$ is an infinitesimal, then so is $f(x)-f(x_{0})$, which is of the same order of infinitesimal as $x-x_{0}$.The idea of infinitesimals served them well but had a little problem in its definition. They were introduced seemed to run against the Archimedean property. The definition of infinitesimals can be made rigorous But, we do not go into it here. However, we can still usefully deal with concepts and notation like:

(a) $f(x)=\mathcal{O}(g(x))$ as $x \rightarrow x_{0}$ if there exists a K such that $|f(x)| \leq K|g(x)|$ for x sufficiently near $x_{0}$.

(b) $f(x)=\mathcal{o}(g(x))$ as $x \rightarrow x_{0}$ if $\lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)}=0$.

Informally, $f(x)=\mathcal{o}(g(x))=0$ means $f(x)$ is of smaller order than $g(x)$ as

$x \rightarrow x_{0}$. In this notation, f is differentiable at $x_{0}$ if there is an l such that

$|f(x)-f(x_{0})-l(x-x_{0})|=\mathcal{o}(|x-x_{0}|)$.

We shall return to this point again. Let us first give examples of derivatives of some functions.

Examples.

(The proof are left as exercises).

(a) $f(x)=x^{n}$, $f^{'}(x_{0})=\lim_{x \rightarrow x_{0}}\frac{x^{n}-{x_{0}}^{n}}{x-x_{0}}=n{x_{0}}^{n-1}$, n a positive integer.

(b) $f(x)=x^{n}$ ($x \neq 0$, where n Is a negative integer), $f^{'}(x)=nx^{n-1}$

(c) $f(x)=e^{x}$, $f^{'}(x)=e^{'}(x)$

(d) $f(x)=a^{x}$, $f^{'}(x)=a^{x}\log[e]{a}$