हिंदी
Tamil Nadu Board of Secondary EducationHSC Science Class 11

Revision: Differential Calculus - Differentiability and Methods of Differentiation Mathematics HSC Science Class 11 Tamil Nadu Board of Secondary Education

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Definitions [1]

Definition: Differentiation

"dy/dx is called the derivative of y with respect to x (which is the rate of change of y with respect to change in x) and the process of finding the derivative is called differentiation."

Theorems and Laws [4]

Prove that the function f given by f(x) = |x − 1|, x ∈ R is not differentiable at x = 1.

Any function will not be differentiable if the left-hand limit and the right-hand limit are not equal.

f(x) = |x − 1|, x ∈ R

f(x) = (x − 1), if x − 1 > 0

= −(x − 1), if x − 1 < 0

At x = 1

f(1) = 1 − 1 = 0

left-side limit:

`lim_(h -> 0^-) (f(1 - h) - f(1))/ -h`

= `lim_(h -> 0^-) (1 - (1 - h) - 0)/ (- h)`

= `lim_(h -> 0^-) (+ h)/(- h)`

= −1

Right-side limit:

= `lim_(h -> 0^+) (f(1 + h) - f(1))/h`

= `lim_(h -> 0^+) ((1 + h) - 1 - 0)/ h`

= `lim_(h -> 0^+) h/h`

= 1

Left-side limit and the right-side limit are not equal.

Hence, f(x) is not differentiable at x = 1.

Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.

Any function will not be differentiable if the left-hand limit and the right-hand limit are not equal.

f(x) = [x], 0 < x < 3

(i) At x = 1

Left-side limit:

`lim_(h -> 0) ([1 - h] - [1])/-h`

= `lim_(h -> 0) (0 - 1)/-h`

= `lim_(h -> 0) 1/h`

= Infinite (∞)

Right-hand limit:

`lim_(h -> 0) ([1 + h] - [1])/h`

= `lim_(h -> 0) (1 - 1)/h`

= 0

Left-side limit and right-side limit are not equal.

Hence, f(x) is not differentiable at x = 1.

(ii) At x = 2

Left-side limit:

`lim_(h -> 0) (f(2 + h) - f(2))/h`

= `lim_(h -> 0) ([2 + h]-2)/h`

= `lim_(h -> 0) (2 -2)/h`

= 0

Right-hand limit:

`lim_(h -> 0) (f(2 - h) - f (2))/h`

= `lim_(h -> 0) ([2 - h] - [2])/-h`

= `lim_(h -> 0) (1 - 2)/-h`

= Infinite (∞)

Left-side limit and right-side limit are not equal.

Hence, f(x) is not differentiable at x = 2.

If y = `[(f(x), g(x), h(x)),(l, m,n),(a,b,c)]`, prove that `dy/dx = |(f'(x), g'(x), h'(x)),(l,m, n),(a,b,c)|`.

y = `|(f(x), g(x), h(x)),(l, m, n),(a, b, c)|`

`dy/dx= |(d/dx (f(x)), d/dx (g(x)), d/dx (h(x))), (l, m, n), (a, b, c)| + |(f(x), g(x), h(x)),(0, 0, 0),(a, b, c)| + |(f(x), g(x), h(x)),(l, m, n),(0, 0, 0)|`

`= |(f'(x), g'(x), h'(x)),(l, m, n),(a, b, c)|`

If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that `[1+ (dy/dx)^2]^(3/2)/((d^2y)/dx^2)` is a constant independent of a and b.

Given, (x – a)2 + (y – b)2 = c2  ...(1)

On differentiating with respect to x,

`=> 2 (x - a) + 2(y - b) dy/dx = 0`

`=> (x - a) + (y - b) dy/dx = 0`  ...(2)

Differentiating again with respect to x,

`1 + dy/dx * dy/dx + (y - b) (d^2 y)/dx^2` = 0

`1 + (dy/dx)^2 + (y - b) (d^2y)/dx^2` = 0

`=> (y - b) = - {(1 + (dy/dx)^2)/((d^2y)/dx^2)}`  ...(3)

Putting the value of (y – b) in (2),

`(x - a) = {(1 + (dy/dx)^2)/((d^2y)/dx^2)}(dy/dx)`   ...(4)

Putting the values ​​of (x − a) and (y − b) from (3) and (4) in (1),

`{1 + (dy/dx)^2}^2/((d^2y)/dx^2)^2 * (dy/dx)^2 + {(1 + (dy/dx)^2)/((d^2y)/dx^2)} = c^2`

On multiplying by `((d^2y)/dx^2)^2`,

`[1 + (dy/dx)^2]^2 (dy/dx)^2 + [1 + (dy/dx)^2]^2 = c^2 ((d^2y)/dx)^2`

`=> [1 + (dy/dx)^2]^2 [(dy/dx)^2 + 1] = c^2 ((d^2y)/dx^2)^2`

`=> {1 + (dy/dx)^2}^3 = c^2 ((d^2y)/dx^2)^2`

On taking the square root,

`therefore {1 + (dy/dx)^2}^(3//2)/((d^2y)/dx^2)` = c  ...(a constant independent of a and b.)

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