Differentiability of a Function and Rate of Change

D. Differentiability

Definition of Tangent

If f is defined on an open interval containing c, and if the limit

The slope of the tangent line to the graph of f at the point (c, f(c)) is also called the slope of the graph of f at x = c.

The above definition of a tangent line to a curve does not cover the possibility of a vertical tangent line. For vertical tangent lines, use the following definition. If f is continuous at c and

then the vertical line x = c, passing through (c, f(c)), is a vertical tangent line to the graph of f. For example, the function shown in the figure has a vertical tangent line at (c, f(c)). If the domain of f is the closed interval [a, b], extend the definition of a vertical tangent to include endpoints by considering one-sided limits (from the right at x = a and from the left at x = b).

The Derivative as a Function

In the preceding section we considered the derivative of a function f at a fixed number a:

Note that alternatively, we can define

Now change the point of view: let the number a vary. If we replace a in Equation (1) by a variable x, we obtain

Given any number x for which this limit exists, we assign to x the number f'(x). So we can regard f' as a new function, called the derivative of f and defined by Equation (2). Geometrically, f'(x) is the slope of the tangent line to the graph of f at (x, f(x)).

The function f' is called the derivative of f because it is obtained from f by this limiting process. The domain of f' is the set {x | f'(x) exists} and may be smaller than the domain of f.

Average and Instantaneous Rate of Change

Suppose y is a function of x, say y = f(x). For a change from x to x + Δx, y changes from f(x) to f(x + Δx). The change in y is Δy = f(x + Δx) - f(x), and the average rate of change of y with respect to x over this interval is

Average rate of change =

As the interval over which we are averaging becomes shorter (that is, as

Thus,

Instantaneous Rate of Change =

Summary

Instantaneous Rate of Change

Suppose f(x) is differentiable at x = x₀. Then the instantaneous rate of change of y = f(x) with respect to x at x₀ is the value of the derivative of f at x₀. That is

Instantaneous Rate of Change = f'(x₀) =

Ex.13 Find the rate at which the function y = x² sin x is changing with respect to x when x =

.

For any x, the instantaneous rate of change in the derivative,

Sol.

The derivative of y = x² sin x is

f'(x) = 2x sin x + x² cos x

Evaluate at x = π:

f'(π) = 2π sin π + π² cos π

Since sin π = 0 and cos π = -1,

f'(π) = 0 + π²(-1) = -π²

The negative sign indicates that when x = π, the function is decreasing at the rate of

Example comparing average and instantaneous rates

Ex.14 Let f(x) = x² - 4x + 7.

(a) Find the instantaneous rate of change of f at x = 3.

(b) Find the average rate of change of f with respect to

x between x = 3 and 5.

Sol.

(a) The derivative of the function is

f'(x) = 2x - 4

Thus, the instantaneous rate of change of f at x = 3 is

f'(3) = 2(3) - 4 = 2

The tangent line at x = 3 has slope 2.

(b) The change in f from x = 3 to x = 5 is

f(5) - f(3) = [5² - 4(5) + 7] - [3² - 4(3) + 7] = 8

Thus, the average rate of change is

The slope of the secant line joining the points (3, f(3)) and (5, f(5)) is 4.

Derivability Over an Interval

Definition : A function f is differentiable at a if f'(a) exists. It is differentiable on an open interval (a, b) if it is differentiable at every number in the interval.

Derivability Over an Interval : f(x) is said to be derivable over an interval if it is derivable at each and every point of the interval. f(x) is said to be derivable over the closed interval [a, b] if :

1. for the endpoints a and b, the one-sided derivatives f'(a⁺) and f'(b^-) exist, and
2. for any point c with a < c < b, the left-hand and right-hand derivatives f'(c^-) and f'(c⁺) exist and are equal.

How Can a Function Fail to Be Differentiable?

There are three common ways a function may fail to be differentiable at a point:

• If the graph has a corner or cusp at the point (for example, y = |x| at x = 0); then the left and right derivatives are different and no tangent exists there.
• If the function is discontinuous at the point (for example, a jump discontinuity), then it cannot be differentiable there.
• If the tangent becomes vertical at the point (a vertical tangent), so that the derivative tends to ±∞.

Right-hand and Left-hand Derivatives

By definition:

f'(a) =

(i) The right-hand derivative of f at x = a, denoted by f'₊(a), is defined by:

f'₊(a) =

(ii) The left-hand derivative of f at x = a, denoted by f'_-(a), is defined by:

f'_-(a) =

We also write f'₊(a) = f'(a⁺) and f'_-(a) = f'(a^-). The derivative f'(a) exists if and only if these one-sided derivatives exist and are equal.

Ex.20 If a function f is defined by f(x) =

show that f is continuous but not derivable at x = 0

Sol.

Compute the right-hand and left-hand limits at 0:

f(0 + 0) =

f(0 - 0) =

Also f(0) = 0. Therefore

f(0 + 0) = f(0 - 0) = f(0) ⇒ f is continuous at x = 0.

Now compute the one-sided derivatives:

f'(0 + 0) =

f'(0 - 0) =

Since f'(0 + 0) ≠ f'(0 - 0), the derivative f'(0) does not exist; thus f is not differentiable at x = 0.

Ex.21 A function f(x) is such that

, if it exists.

Sol.

Given that =

Ex.22 Let f be differentiable at x = a and let f(a) ≠ 0. Evaluate

Sol.

l =

l =

Ex.23 Let f : R → R satisfying

then show f(x) is differentiable at x = 0.

Sol.

Since,

Now,

From (ii) and (iii), we get f'(0) = 0. Therefore f(x) is differentiable at x = 0.

F. Operations on Differentiable Functions

1. If f(x) and g(x) are differentiable at x = a then the functions f(x) + g(x), f(x) - g(x), and f(x)·g(x) are differentiable at x = a. If g(a) ≠ 0 then f(x)/g(x) is differentiable at x = a.

If f and g are differentiable functions, then their product fg is differentiable. Proof:

Let a be a number in the domain of fg. By the definition of the product of two functions we have

(fg)(a) = f(a) g(a)

(fg)(a + t) = f(a + t) g(a + t)

Hence

(fg)'(a) =

The following algebraic manipulation will enable us to put the above fraction into a form in which we can see its limit:

f(a + t) g(a + t) - f(a) g(a) = f(a + t) g(a + t) - f(a) g(a + t) + f(a) g(a + t) - f(a) g(a)

= [f(a + t) - f(a)] g(a + t) + [g(a + t) - g(a)] f(a)

Thus

(fg)'(a) =

The limit of a sum is the sum of the limits. Moreover, f'(a) and g'(a) exist by hypothesis. Because g is differentiable at a, it is continuous there, so

(fg)'(a) =

= f'(a) g(a) + g'(a) f(a) = (f' g + g' f)(a).

2. If f(x) is differentiable at x = a and g(x) is not differentiable at x = a, then the product F(x) = f(x)·g(x) can still be differentiable at x = a. Example: f(x) = x and g(x) =

3. If f(x) and g(x) are both not differentiable at x = a, the product F(x) = f(x)·g(x) can still be differentiable at x = a. Example: f(x) =

4. If f(x) and g(x) are both non-differentiable at x = a then the sum F(x) = f(x) + g(x) may be differentiable. Example: f(x) =

5. If f(x) is differentiable at x = a, then f'(x) is not necessarily continuous at x = a. Example: f(x) =

G. Functional Equations Involving Differentiability

Ex.24 Let f(xy) = x f(y) + y f(x) for all x,

and f(x) be differentiable in (0, ∞) then determine f(x).

Given f(xy)= x f(y) + y f(x)

Sol. Replacing x by 1 and y by x then we get x f(1) = 0

On integrating with respect to x and taking limits from 1 to x, we obtain

f(x)/x - f(1)/1 = f'(1) (ln x - ln 1)

Since f(1) = 0, we get f(x) = f'(1) (x ln x).

Alternative Method :

Differentiate both sides of f(xy) = x f(y) + y f(x) with respect to x, treating y as constant. Then f'(xy)·y = f(y) + y f'(x).

Putting y = 1 gives a relation that leads, on integrating and using f(1) = 0, to

f(x) = f'(1) (x ln x).

Ex.25 If

and f'(1) = e, determine f(x).

Sol.

Given e^-xy f(xy) = e^-x f(x) + e^-y f(y) ....(1)

Putting x = y = 1 in (1), we get f(1) = 0 ...(2)

On integrating we obtain e^-x f(x) = ln x + C. Using x = 1 and f(1) = 0, we find C = 0.

Therefore f(x) = e^x ln x.

Ex.26 Let f be a function such that f(x + f(y)) = f(f(x)) + f(y) x, y

where ε > 0, then determine f''(x) and f(x).

Sol.

Given f(x + f(y)) = f(f(x)) + f(y) .....(1)

Put x = y = 0 in (1): f(0 + f(0)) = f(f(0)) + f(0) ⇒ f(f(0)) = f(f(0)) + f(0)

Thus f(0) = 0 ...(2)

Integrating both sides with limits 0 to x yields f(x) = x. Therefore f'(x) = 1 and f''(x) = 0.

Optional brief summary

Differentiability is the existence of a finite tangent slope (derivative) at a point. The derivative represents the instantaneous rate of change. Functions fail to be differentiable at corners, discontinuities, or vertical tangents. One-sided derivatives decide differentiability at a point. Differentiable functions enjoy standard algebraic rules (sum, product, quotient) and these rules are useful in solving functional equations that involve differentiable functions.