Differentiability of a Function and Rate of Change | Mathematics (Maths) Class 12 - JEE PDF Download

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D. Differentiability 

Definition of Tangent : If f is defined on an open interval containing c, and if the limit

 = m exists, then the line passing through (c, f(c)) with slope m is the tangent line to the graph of f at the point (c, f(c)).

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, you can 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 Figure has a vertical tangent line at (c, f(c)). If the domain of f is the closed interval [a, b], then you can extend the definition of a vertical tangent line to include the endpoints by considering continuity and limits from the right (for x = a) and from the left (for x = b).

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

  .....(1)

Note that alternatively, we can define

  provided the limit exists.

Here we change our point of view and let the number a vary. If we replace a in Equation 1 by a variable x,

we obtain   ...(2)

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. We know that the value of f'(x), can be interpreted geometrically as the slope of the tangent line to the graph of f at the point (x, f(x)).

The function f' is called the derivative of f because it has been "derived" from f by the limiting operation in Equation 2. 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). Corresponding to a change from x to x + Δx, the variable 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 is

Average rate of change =

As the interval over which we are averaging becomes shorter (that is, as  ), the average rate of change approaches what we would intuitively call the instantaneous rate of change of y with respect to x, and the difference quotient approaches the derivative   Thus, we have

Instantaneous Rate of Change =

To summarize :

Instantaneous Rate of Change 

Suppopse f(x) is differentiable at x = x₀. Then the instantaneous rate of cange 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.

 = 2π sin π + π² cos π = 2π(0) + π² (-1) = -π²

The negative sign indicates that when x = π , the function is decreasing at the rate of  units of y for each one-unit increase in x.

Let us consider an example comparing the average rate of change and the instantaneous rate of change.

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, as shown in the figure

(b) The (average) rate of change from x = 3 to x = 5 is found by dividing the change  in f by the change in x. 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 is 4, as shown in the figure.

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)   [or  ] 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 & every point of the interval.  f(x) is said to be derivable over the closed interval [a, b] if :

(i) for the points a and b, f '(a+) & f '(b -) exist &
(ii) for any point c such that a < c < b, f '(c+) & f'(c -) exist & are equal .

How Can a Function Fail to Be Differentiable ? 

We see that the function y = |x| is not differentiable at 0 and Figure shows that its graph changes direction abruptly when x = 0. In general, if the graph of a function f has a "corner" or "kink" in it, then the graph of f has no tangent at this point and f is not differentiable there. [In trying to compute f '(a), we find that the left and right limits are different.]

There is another way for a function not to have a derivative. If f is discontinuous at a, then f is not differentiable at a. So at any discontinuity (for instance, a jump discontinuity), f fails to be differentiable.

A third possibility is that the curve has a vertical tangent line when at x = a,

This means that the tangent lines become steeper and steeper as x → a. Figure (a, b, c) illustrates the three posibilities that we have discussed.

Right hand & 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) =  , provided the limit exists & is finite.

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

f ' -(a) =   , Provided the limit exists & is finite. We also write f '₊(a) = f '(a⁺) & f '-(a) = f '(a^-) .

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. We have f(0 + 0) =  = 0

f(0 - 0) =  = 0

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

Again f'(0 + 0) =  = 1

f'(0 - 0) =  = 0

Since f'(0 + 0)  f'(0 - 0), the derivative of f(x) at x = 0 does not exist.

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) ¹ 0. Evaluate .

 Sol. l =   (1^∞ form)

l =   (put n = 1/h)

Ex.23 Let f : R → R satisfying  then show f(x) is differentiable at x = 0.

Sol. Since,  f(0) = 0    ...(i)

   ....(ii) {f(0) = 0 from (i)}

Now,  → 0 ...(iii) {using Cauchy-Squeeze theorem}

 from (ii) and (iii) , we get f'(0) = 0. i.e. f(x) is differentiable at x = 0.

F. Operation on Differentiable Functions

1. If f(x) & g(x) are derivable at x = a then the functions f(x) + g(x), f(x) - g(x), f(x). g(x) will also be derivable at x = a & if g (a)  0 then the function f(x)/g(x) will also be derivable at x = a.

If f and g are differentiable functions, then prove that their product fg is differentiable.

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 what the limit is:

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 of products is the sum of the products of the limits. Moreover, f'(a) and g'(a) exist by hypothesis. Finally, since g is differentiable at a, it is continuous there ; and so  = f(a). We conclude that

(fg)'(a) =

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

2. If f(x) is differentiable at x = a & g(x) is not differentiable at x = a , then the product function F(x) = f(x) . g(x) can still be differentiable at x = a e.g. f(x) = x  and g(x) =  .

3. If f(x) & g(x) both are not differentiable at x = a then the product function ;

F(x) = f(x) . g(x) can still be differentiable at x = a e.g. f(x) =  & g(x) =

4. If f(x) & g(x) both are non-deri. at x = a then the sum function F(x) = f(x) + g(x) may be a differentiable function . e.g. f(x) =  & g(x) = - . 

5. If f(x) is derivable at x = a ⇒ f '(x) is continuous at x = a.

e.g. f(x) =

G. Functional Equations 

Ex.24 Let f(xy) = xf(y) + yf(x) for all x,  and f(x) be differentiable in (0, ∞) then determine f(x).

Given f(xy)= xf(y) + yf(x) 

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

On integrating w.r.t.x and taking limit 1 to x then f(x)/x - f(1)/1 = f'(1) (ln x – ln 1)

∴  f(1) = 0) ∴  f(x) = f'(1) (x ln x)

Alternative Method : 

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

Differentiating both sides w.r.t.x treating y as constant, f'(xy) . y = f(y) + yf'(x)

Putting y = x and x = 1, then

f'(xy). x = f(x) + xf'(x)

Integrating both sides w.r.t.x taking limit 1 to x,

Hence, f(x) =- f'(1)(x ln x).

Ex.25 If  and f'(1) = e, determine f(x).

Sol.

Given e^–xy f(xy) = e^–xf(x) + e^–yf(y) ....(1)

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

On integrating we have e^–xf(x) = ln x + c at x = 1, c = 0

∴  f(x) = ex 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), then f(0 + f(0)) = f(f(0)) + f(0)  ⇒    f(f(0)) = f(f(0)) + f(0)

∴ f(0) = 0 ...(2)

Integrating both sides with limites 0 to x then f(x) = x   ∴ f'(x) = 1.