Differentiation

Introduction

In the previous chapter we discussed about the continuity of a function that is a function is continuous at point $x = a$ , then its graph is an unbroken curve at $x = a$ . In this chapter we deal with yet another concept, differentiability of function. Before discussion we first define a few useful terms.

Rate of Change

Let $y = f(x)$ be a function of $x$ . Putting the values of ' $x$ ' in this relation, we obtain the corresponding values of ' $y$ '. Suppose we start putting some values of ' $x$ ' in increasing order. The respective values of ' $y$ ' that we obtain may turn out to be in increasing order, or in decreasing order, or they may remain constant, or they may even have a mixed trend, depending upon the type of function. Let us take two values of x : $x_{1}$ and $x_{2}\left( x_{1} < x_{2} \right)$ , so that, $y_{1} = f\left( x_{1} \right)$ and $y_{2} = f\left( x_{2} \right)$ .

Then, the quantity $\left( \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \right)$ will tell us the average rate of change of $y$ w.r.t. $x$ in the interval $\left\lbrack x_{1},x_{2} \right\rbrack$ .

Let $y_{2} > y_{1} \Rightarrow \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ is positive ⇒ Function is increasing on an average.

Let $y_{2} < y_{1} \Rightarrow \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ is negative ⇒ Function is decreasing on an average.

Let $y_{2} = y_{1} \Rightarrow \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ is zero ⇒ Function is constant on an average.

As we see, if $x_{1}$ and $x_{2}$ are sufficiently far apart, the quantity $\left( \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \right)$ can not given the exact idea of the variation of $y$ w.r.t $x$ in the interval $\left\lbrack x_{1},x_{2} \right\rbrack$ . It just provides an overall information. For example, if $y_{2} = y_{1}$ , it does not necessarily means that $y$ is the same for all x in the interval $\left\lbrack x_{1},x_{2} \right\rbrack$ . Thus, in order to obtain a "sufficiently accurate information", we have to choose $x_{1}$ and $x_{2}$ "sufficiently close" to each other. This 'sufficiently close' is the key word here. To know the rate of change y w.r.t x at $x = x_{1}$ , we take $x_{2}$ very near to

$x_{1}$ (as much as possible) i.e., ' $x_{2}$ tends to $x_{1}$ ' and then calculate $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ . In the limiting case.

We say that $x_{2}$ nearly coincides with $x_{1}$ and represent it as $x_{2} \rightarrow x_{1}$ . We use the notation $\left. \ \frac{dy}{dx} \right|_{x = x_{1}}$ for $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ as $X_{2} \rightarrow X_{1}$ .

' $dx$ ' means small change in $x$ (near $x = x_{1}$ ) and ' $dy$ ' means the corresponding change in $y$ .

We call $\frac{dy}{dx}$ the derivative or the differential coefficient of y w.r.t. x .

Differentiability of a Function at a given Point

Let $f(x)$ be a real valued function defined on an open interval (a,b) where $c \in (a,b)$ . Then $f(x)$ is said to be differentiable or derivable at $x = c$ ,

iff $\lim_{x \rightarrow c}\mspace{2mu}\frac{f(x) - f(c)}{(x - c)}$ exists finitely.

This limit is called the derivative or differential coefficient of the function

$f(x)$ at $x = c$ , and is denoted by $f^{'}(C)$ or $Df(C)$ or $\frac{d}{dx}.(f(x))_{x = c}$

⇔ Thus, $f(x)$ is differentiable at $x = c$ if $\lim_{x \rightarrow c}\mspace{2mu}\frac{f(x) - f(c)}{(x - c)}$ exists finitely

$\Rightarrow \lim_{x \rightarrow c^{-}}\mspace{2mu}\frac{f(x) - f(c)}{(x - c)} = \lim_{x \rightarrow c^{+}}\mspace{2mu}\frac{f(x) - f(c)}{(x - c)} \Rightarrow \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(c - h) - f(c)}{( - h)} = \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(c + h) - f(c)}{(h)}$

Here $\lim_{x \rightarrow c^{-}}\mspace{2mu}\frac{f(x) - f(c)}{(x - c)} = \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(c - h) - f(c)}{( - h)}$ is called the left hand derivative of $f(x)$ at $x = c$ and is denoted by $f^{'}\left( c^{-} \right)$ or $LF^{'}(C)$ . While, $\lim_{x \rightarrow c^{+}}\mspace{2mu}\frac{f(x) - f(c)}{(x - c)} = \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(c + h) - f(c)}{(h)}$ is called the right hand derivative of $f(x)$ at $x = c$ and is denoted by $f^{'}\left( c^{+} \right)$ or Rf ' (C).

Thus $f(x)$ is differentiable at $x = c$ , if $Lf^{'}(C) = Rf^{'}(C)$ .

If $Lf^{'}(C) \neq Rf^{'}(C)$ we say that $f(x)$ is not differentiable at $x = c$ .

Derivative of $\mathbf{f(x)}$ from the First Principle

If $f(x)$ is derivable function then $\lim_{h \rightarrow 0}\mspace{2mu}\frac{f(x + h) - f(x)}{h} = f^{'}(x) = \frac{dy}{dx}$ is called the first principle of calculating the differentiability coefficient of any given function.

Illustration:

Differentiate $\mathbf{\sin}\mathbf{x}^{\mathbf{2}}$ from first principles

Solution:

$\begin{array}{cl} \frac{d}{d x} s i n x^{2} = lim_{h \to 0} \frac{s i n {(x + h)}^{2} - s i n x^{2}}{h} = lim_{h \to 0} {\frac{s i n {(x + h)}^{2} - s i n x^{2}}{{(x + h)}^{2} - x^{2}} \cdot \frac{{(x + h)}^{2} - x^{2}}{h}} \\ = lim_{h \to 0} {\frac{2 c o s \frac{1}{2} {{(x + h)}^{2} + x^{2}} s i n \frac{1}{2} {{(x + h)}^{2} - x^{2}}}{2 h x + h^{2}} \frac{2 h x + h^{2}}{h}} \\ = lim_{h \to 0} {c o s \frac{1}{2} {{(x + h)}^{2} + x^{2}} \frac{s i n \frac{1}{2} (2 x h + h^{2})}{\frac{1}{2} (2 h x + h^{2})} (2 x + h)} = (c o s x^{2}) 2 x \\ Because as h \to 0, \frac{s i n \frac{1}{2} (2 h x + h^{2})}{\frac{1}{2} (2 h x + h^{2})} \to 1; 2 x + h \to 2 x \\ and c o s \frac{1}{2} {{(x + h)}^{2} + x^{2}} \to c o s \frac{1}{2} {x^{2} + x^{2}} \\ Therefore \frac{d}{d x} s i n x^{2} = 2 x c o s x^{2} . \end{array}$

Geometrical Meaning of Differentiability

Consider the function $f(x)$ defined on an open interval ( $a,b$ ) let $P(c,f(C))$ be a point on the curve $y = f(x)$ and let $Q((c - h),f(c - h))$ and $R((c + h),f(c + h))$ be two neighbouring points on the left hand side and right hand side respectively of point $P$ as shown in figure.

Then slope of chord $PQ = \frac{f(c - h) - f(c)}{(c - h) - (c)} = \frac{f(c - h) - f(c)}{- h}$

and, slope of chord PR $= \frac{f(c + h) - f(c)}{(c + h) - (c)} = \frac{f(c + h) - f(c)}{h}$

We know that tangent to a curve at a point $P$ (say) is the limiting position of chord $PQ$ when $Q \rightarrow P$ . Therefore, as $h \rightarrow 0$ , points $Q$ and $R$ both tends to $P$ from left hand and right hand sides respectively. Consequently, chord PQ and PR become tangents at point P .

Thus, $\lim_{h \rightarrow 0}\mspace{2mu}\frac{f(c - h) - f(c)}{( - h)} = \lim_{h \rightarrow 0}\mspace{2mu}$ (slope of chord $\left. \ PQ \right) = \lim_{Q \rightarrow p}\mspace{2mu}$ (slope of chord $PQ$ )

⇒ Slope of the tangent at point P , which is limiting position of the chords

drawn on the left hand side of point $P$ .

and, $= \lim_{h \to 0} \frac{f (c + h) - f (c)}{h} = \lim_{h \to 0}$ (slope of chord PR) $= \lim_{R \to P}$ (slope of chord PR)

Slope of the tangent at point P , which is limiting position of the chords drawn on the left hand side of point P .

Now, $F(x)$ is differentiable at $x = c \Leftrightarrow \lim_{h \to 0} \frac{f (c - h) - f (c)}{(- h)} = \lim_{h \to 0} \frac{f (c + h) - f (c)}{h}$

⇔ There is a unique tangent at point P .

Thus , $f(x)$ is differentiable at point $p$ , if there exists a unique tangent at point $P$ . In other words, $f(x)$ is differentiable at a point P if the curve does not have P as a corner point. "i.e, the function is not differentiable at those points on which function has jumps (or holes) and sharp edges.

Reasons for Non-Differentiability

If a function $y = f(x)$ is non-differentiable at $x = a$ then it may be due to one of following reasons:

(i) $\ f(x)$ is discontinuous at $x = a$

(ii) $\ f(x)$ tends to become parallel to $y$ -axis in the neighbourhood of $x = a$ .

(iii) $\ f(x)$ has a corner point (or a sharp turn) at $x = a$

Illustration:

Find the values of $\mathbf{x}$ for which the function $f(x) = |cosecx - 2|$ is continuous but not differentiable. Since $f(x)$ is a periodic function with period $= 2\pi$ so, we analyse the function in the interval ( $- \pi,\pi$ ) then generalize the result.

Solution:

The graph of the function $f(x)$ has a corner at $x = \alpha\&\beta$ which are the roots of where $x \in \lbrack 0,2\pi\rbrack$ .

The equation $cosecx = 2$

$\Rightarrow sinx = 1/2 = sin\frac{\pi}{6} \Rightarrow x = n\pi + ( - 1)^{n}\frac{\pi}{6}n \in I$

Illustration:

$L e t f (x) = | x - 1 | + | x + 1 |$ . Discuss the continuity and differentiability of the function.

Solution:

Here $f(x) = |x - 1| + |x + 1| \Rightarrow f(x)\left\{ \begin{matrix} 2x,\text{~when~} & x > 1 \\ 2,\text{~when~} - 1 \leq & x \leq 1 \\ - 2x,\text{~when~} & x < - 1 \end{matrix} \right.\$

Graphical Solution :

The Graph of the function is shown alongside,

From the graph it is clear that the function is continuous at all real $x$ , also differentiable at all real $x$ except at $x = \pm 1$ ; Since sharp edges at $x = - 1$ and $x = 1$

At $x = 1$ we see that the slope from the right i.e., RHD =2, while slope from the left i.e., $LHD = 0$ . Similarly, at $x = - 1$ it is

clear that RHD $= 0$ , while $LHD = - 2$

Analytical Method:

In this method, first of all, we differentiate the function and on the derivative equality sigh should be removed from doubtful points.

Here $f^{'}(x) = \left\{ \begin{matrix} - 2 & x < - 1 \\ 0 & - 1 < x < 1 \\ 2 & x > 1 \end{matrix} \right.\$

Now, at $x = 1,f^{'}\left( 1^{+} \right) = 2$ while $f^{'}\left( 1^{-} \right) = 0$ and At $x = 1,f^{'}\left( - 1^{+} \right) = 0$ while $f^{'}\left( - 1^{-} \right) = - 2$

Thus $f(x)$ is not differentiable at $x = \pm 1$ .

Note: We should not use this method when function is discontinuous.

Illustration:

If the derivative of the function $f(x) = \left\{ \begin{matrix} ax^{2} + b\ ,x < - 1 \\ bx^{2} + ax + 4,x \geq - 1 \end{matrix} \right.\$ is everywhere continuous and differentiable at $x = 1$ then find $a$ and $b$ .

Solution:

$f(x) = \left\{ \begin{matrix} ax^{2} + b & ,x < - 1 \\ bx^{2} + ax + 4, & x \geq - 1 \end{matrix}\ \therefore f^{'}(x) = \left\{ \begin{matrix} 2ax\ & x < - 1 \\ 2bx + a, & x \geq - 1 \end{matrix} \right.\ \right.\$

To find $a,b$ we must have two equations in $a,b$

Since $f(x)$ is differentiable, it must be continuous at $x = - 1$ .

$\therefore R = L = V$ at $x = - 1$ for $f(x) \Rightarrow b - a + 4 = a + b$

$\therefore 2a = 4$ i.e., $a = 2$

Again $f^{'}(x)$ is continuous, it must be continuous at $x = - 1$ .

$\therefore R = L = V$ at $x = - 1$ for $f^{'}(x)$

$- 2b + a = - 2a$ . Putting $a = 2$ , we get $- 2b + 2 = - 4$

$\therefore 2\text{ }b = 6$ or $b = 3$ .

Differentiation

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