In the previous section, we completed a discussion on continuous functions. In this section, we will see differentiability.
We have seen derivatives in class 11 (Details here). Let us recall some basic details. It can be written in 8 steps:
1. If f is a real function and c a point in it's domain, then the derivative of f at c is given by:
$f'(c) = \lim_{h\rightarrow 0} \left[\frac{f(c+h) - f(c)}{h} \right]$
2. We can find the derivative at a particular point.
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The derivative at a particular point c is denoted as f'(c).
3. We can also find the general form of the derivative.
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The general form is denoted as f'(x).
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In the general form, we can substitute c in the place of x. This will give the actual derivative at c.
4. f'(c) is the "slope of the tangent" of f at the point c.
5. f'(c) is also the "rate of change" of the output.
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For example, consider the function in which time (t) is the input and velocity (v) is the output. Then the derivative at t = 8 s will give the rate of change of velocity at the instant when the stop-watch reading is 8 s.
6. The process of finding the derivative of a function is called differentiation.
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Some text books use the following phrase:
Differentiate f(x) with respect to x.
♦ This means: we are being asked to find f'(x).
7. We saw three rules related to algebra of derivatives:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(u \pm v)'} & {~=~} &{u' \pm v'} \\
{~\color{magenta} 2 } &{{}} &{(uv)'} & {~=~} &{u' v + u v'~\text{(Leibnitz rule or Product rule)}} \\
{~\color{magenta} 3 } &{{}} &{\left(\frac{u}{v} \right)'} & {~=~} &{\frac{u' v \,-\, u v' }{v^2}~\text{Wherever}~v \ne 0~\text{(Quotient rule)}} \\
\end{array}$
8. We saw some standard results also:
$\begin{array}{ll} {~\color{magenta} 1 } &{\text{If}} &{f(x)} & {~=~} &{x^n} \\
{~\color{magenta} {} } &{\text{ Then}} &{f'(x)} & {~=~} &{n x^{n-1}} \\
{~\color{magenta} {} } &{{}} &{{}} {} &{{}} \\
{~\color{magenta} 2 } &{\text{If}} &{f(x)} & {~=~} &{\sin x} \\
{~\color{magenta} {} } &{\text{ Then}} &{f'(x)} & {~=~} &{\cos x} \\
{~\color{magenta} {} } &{{}} &{{}} {} &{{}} \\
{~\color{magenta} 3 } &{\text{If}} &{f(x)} & {~=~} &{\cos x} \\
{~\color{magenta} {} } &{\text{ Then}} &{f'(x)} & {~=~} &{- \sin x} \\
{~\color{magenta} {} } &{{}} &{{}} {} &{{}} \\
{~\color{magenta} 4 } &{\text{If}} &{f(x)} & {~=~} &{\tan x} \\
{~\color{magenta} {} } &{\text{ Then}} &{f'(x)} & {~=~} &{\sec^2 x} \\
\end{array}$
Now we will derive a more convenient expression for derivative. It can be done in 5 steps:
1. Consider the definition of derivative:
$f'(x)~=~\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}$
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We obtained this expression using fig.13.31 in section 13.13.
2. The above expression can be written in another convenient form by using fig.21.14 below:
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| Fig.21.14 |
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Point P has:
♦ x-coordinate c
♦ y-coordinate f(c)
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Point Q has:
♦ x-coordinate x
♦ y-coordinate f(x)
3. So base of the triangle PQR = PR = (x-c)
Also, altitude = QR = [f(x) − f(c)]
4. Then slope of the line PQ = $\frac{f(x) - f(c)}{x - c}$
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We want PQ to be the tangent at P. So we want Q to be very close to P.
That is., we want x to be very close to c. We write this as x→c.
5. When x is very close to c, the slope of PQ is given by:
$\lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$
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The slope in such a situation is:
the slope of tangent at P.
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Slope of tangent at P is the derivative at c.
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So we can write:
Derivative at c = $f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$
Now we will see the "importance of continuity" in the process of differentiation. It can be written in steps:
1. We saw that, the derivative at c is given by:
$f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$
2. The above expression can be rearranged as:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{f'(c)} & {~=~} &{\lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{\lim_{x\rightarrow c}[f(x) - f(c)]}{\lim_{x\rightarrow c}[x - c]}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{\lim_{x\rightarrow c}[f(x)]~-~\lim_{x\rightarrow c}[f(c)]}{\lim_{x\rightarrow c}[x - c]}} \\
\end{array}$
3. Consider the first term in the numerator:
$\lim_{x\rightarrow c}[f(x)]$
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It shows that, limit at c must exist.
4. We know that:
If limit at c is to exist, both left side and right side limits at c must exist, and they must be the same.
5. We have seen several examples where limit at a particular point does not exist. To cite one example, see fig.21.6 in section 21.2.
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In this example, we cannot find the derivative at x = 1 because, $\lim_{x\rightarrow 1}[f(x)]$ does not exist.
6. While discussing the topic of continuity, we saw that:
If at a point c, both left side and right side limits exist and if they are equal, then the function is continuous at c.
7. Based on this condition for continuity, we can write the condition for differentiability.
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We can write:
If we want to find the derivative of a function f at the point c, then f must be continuous at c.
8. If it is possible to find the derivative of a function f at the point c, then we say that, f is differentiable at c.
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So we can write:
For differentiability at c, "continuity at c" is essential.
9. We have seen continuous functions. Such functions are continuous at every point in the domain.
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So we can write:
If the domain of a continuous function f, is [a,b], then f is differentiable at every point in [a,b]
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As in the case of continuity,
♦ At the point a, we need to consider only the right side limit.
♦ At the point b, we need to consider only the left side limit.
Now we will see a theorem which gives the relation between continuity and differentiability.
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The theorem states that:
If a function f is differentiable at a point c, then the function is continuous at c.
The proof can be written in 3 steps:
1. Consider the expression for derivative that we obtained above:
$f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$
2. Now consider the case when x ≠ c.
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We can write f(x) - f(c) as:
$f(x) - f(c) = {\frac{f(x) - f(c)}{x-c}}.(x-c)$
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Taking limits on both sides, we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\lim_{x\rightarrow c} \left[f(x) - f(c) \right]} & {~=~} &{\lim_{x\rightarrow c} \left[{\frac{f(x) - f(c)}{x-c}}.(x-c) \right]} \\
{~\color{magenta} 2 } &{\implies} &{\lim_{x\rightarrow c} \left[f(x) \right] - \lim_{x\rightarrow c} \left[f(c) \right]} & {~=~} &{\lim_{x\rightarrow c} \left[{\frac{f(x) - f(c)}{x-c}} \right].\lim_{x\rightarrow c} \left[(x-c) \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{f'(c) . 0} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{0} \\
{~\color{magenta} 5 } &{\implies} &{\lim_{x\rightarrow c} \left[f(x) \right]} & {~=~} &{\lim_{x\rightarrow c} \left[f(c) \right]} \\
{~\color{magenta} 6 } &{\implies} &{\lim_{x\rightarrow c} f(x)} & {~=~} &{f(c)} \\
\end{array}$
3. We want to prove that, f is continuous at c. Recall that, we need to check two conditions to prove continuity.
(i) Limit exists at c.
(ii) $\lim_{x\rightarrow c} f(x) = f(c)$
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In the theorem, it is given that, f is differentiable at c. So it is
obvious that, limit at c exists. Thus condition (i) is satisfied.
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From (3) we have: $\lim_{x\rightarrow c} f(x) = f(c)$.
Thus condition (ii) is also satisfied.
◼ Hence it is proved that, f is continuous at c.
◼ As a corollary of the above theorem, we can write:
Every differentiable function is continuous.
(Corollary of a theorem, is a fact, which results directly from that theorem. The dictionary meaning can be seen here)
The converse of the above corollary is not true. That is., every continuous function is not differentiable. Let us see an example. It can be written in steps:
1. Consider the function f(x) = |x|.
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We know that, it is a continuous function. We want to prove that, it is not a differentiable function.
2. Let us try to find the derivative of this function at x = 0
3. We have the general equation:
$f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$
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So the derivative at x = 0 is given by:
$f'(0) = \lim_{x\rightarrow 0} \left[\frac{|x| - f(0)}{x - 0} \right]$
4. Let us write the left side limit at x = 0:
$\lim_{x\rightarrow 0^{-}} \left[\frac{|x| - f(0)}{x - 0} \right]$
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As x approaches zero from the left side, the input will be a −ve value. Let it be −h.
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So the left side limit will be:
$\frac{|-h| - |0|}{(-h) - 0} ~=~\frac{h}{(-h)}~=~-1$
5. Let us write the right side limit at x = 0:
$\lim_{x\rightarrow 0^{+}} \left[\frac{|x| - f(0)}{x - 0} \right]$
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As x approaches zero from the right side, the input will be a +ve value. Let it be h.
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So the right side limit will be:
$\frac{|h| - |0|}{h - 0}~=~\frac{h}{h}~=~1$
6. From (4) and (5), we see that:
Left side and right side limits are not the same. So the required limit does not exist at x = 0.
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Therefore, the derivative at x = 0 does not exist.
7. Based on this example, we can write:
All continuous functions are not differentiable.
In the next section, we will see derivatives of composite functions.
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