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 {xf'(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 = x_{0}. Then the instantaneous rate of cange of y = f(x) with respect to x at x_{0} is the value of the derivative of f at x_{0}. That is
Instantaneous Rate of Change = f'(x_{0}) =
Ex.13 Find the rate at which the function y = x^{2} 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 + π^{2} cos π = 2π(0) + π^{2} (1) = π^{2}
The negative sign indicates that when x = π , the function is decreasing at the rate of units of y for each oneunit 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) = x^{2}  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) = [5^{2} – 4(5) + 7] – [3^{2} – 4(3) + 7] = 8
Thus, the average rate of change is
The slope of the secant line is 4, as shown in the figure.
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 onesided 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 CauchySqueeze 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 nonderi. 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^{–x}f(x) + e^{–y}f(y) ....(1)
Putting x = y = 1 in (1), we get f(1) = 0 ...(2)
On integrating we have e^{–x}f(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.
204 videos288 docs139 tests

1. What is the difference between derivability and differentiability? 
2. How are derivability and differentiability related to the rate of change of a function? 
3. Can a function be differentiable but not derivable? 
4. What does it mean for a function to be differentiable over an interval? 
5. Can a function be differentiable over an interval but not continuous? 

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