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Revision: Calculus >> Continuity and Differentiability Maths Commerce (English Medium) Class 12 CBSE

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Definitions [9]

Definition: Continuous Function

A function f(x) is said to be continuous at a point x = a, if the following three conditions are satisfied

  1. f is defined at every point on an open interval containing a.
  2. \[\lim_{x\to a}f\left(x\right)\] exists.
  3. \[\lim_{x\to a}f\left(x\right)=f\left(a\right)\].
Definition: Equation of Continuity

The principle which states that for a non-viscous liquid in streamline flow passing through a tube of varying cross-section, the product of the area of cross-section and the velocity of flow remains constant at every point is called the Equation of Continuity.

Definition: Discontinuous Function

A function f(x) is said to be discontinuous at x = a if it is not continuous at x = a, i.e.

  1. \[\lim_{x\to a}f\left(a\right)\] does not exist.
  2. The left-hand limit and the right-hand limit are not equal.
  3. \[\lim_{x\to a}f\left(x\right)\neq f\left(a\right)\].
Definition: Removable Discontinuity

If \[\lim_{x\to a^{-}}f\left(x\right)=\lim_{x\to a^{+}}f\left(x\right)\neq f\left(a\right),\] then f(x) is said to be removable discontinuous.

Definition: Non Removable Discontinuity

If \[\lim_{x\to a^{+}}f\left(x\right)\neq\lim_{x\to a^{-}}f\left(x\right),\] then f(x) is said to be non-removable discontinuous.

Definition: Domain of Continuity

The set of all points where a function is continuous is called its domain of continuity.

Definition: Differentiability

A function f(x) is said to be differentiable at x = a if Rf'(a) and Lf'(a) both exist and are equal; it is said to be non-differentiable. 

Definition: Continuity at a Point

Let f(x) be a real function and a be a point in its domain.

A function f is continuous at x = a iff all three conditions hold:

  • f(a) is defined
  • \[\lim_{x\to a}f(x)\] exists
  • \[\lim_{x\to a}f(x)\] = f(a)

\[\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)\]

Definition: Continuity in an Interval

For the open interval:

A function f is said to be continuous on an open interval (a, b) if it is continuous at every point in the interval. 

For a closed interval:

A function f is said to be continuous on the closed interval [a,b] iff:

  1. f is continuous at every point of (a,b)

  2. f is right continuous at a

    \[\lim_{x\to a^+}f(x)=f(a)\]
  3. f is left continuous at b

    \[\lim_{x\to b^-}f(x)=f(b)\]

Formulae [14]

Formula: Equation of Continuity

For a non-viscous liquid in streamline flow passing through a tube of varying cross-section:

av = constant

or equivalently:

a ∝ \[\frac {1}{v}\]

where: 

  • a = area of cross-section of the tube
  • v = velocity of flow of the liquid
Formula: Derivative of Composite Functions
Function Derivative
[f(x)]ⁿ n[f(x)]ⁿ⁻¹ · f′(x)
\[\sqrt{\mathrm{f}(x)}\] \[\frac{1}{2\sqrt{\mathrm{f}(x)}}\cdot\mathrm{f}^{\prime}(x)\]
\[\frac{1}{\mathrm{f}(x)}\] \[-\frac{1}{\left[\mathrm{f}(x)\right]^{2}}\cdot\mathrm{f}^{\prime}(x)\]
sin(f(x)) cos(f(x)) · f′(x)
cos(f(x)) −sin(f(x)) · f′(x)
tan(f(x)) sec²(f(x)) · f′(x)
cot(f(x)) −cosec²(f(x)) · f′(x)
sec(f(x)) sec(f(x)) tan(f(x)) · f′(x)
cosec(f(x)) −cosec(f(x)) cot(f(x)) · f′(x)
\[\mathbf{a}^{\mathbf{f}(x)}\] \[a^{f(x)}\log a\cdot f^{\prime}(x)\]
\[\mathrm{e}^{\mathrm{f}(x)}\] \[\mathrm{e}^{\mathrm{f}(x)\cdot\mathrm{f}^{\prime}(x)}\]
log(f(x)) \[\frac{1}{\mathrm{f}(x)}\cdot\mathrm{f}^{\prime}(x)\]
logₐ(f(x)) \[\frac{1}{\mathrm{f}(x)\mathrm{loga}}\cdot\mathrm{f}^{\prime}(x)\]
Formula: Derivative of Inverse Functions
Function Derivative Condition
sin⁻¹x \[\frac{1}{\sqrt{1-x^{2}}}\] |x| < 1
sin⁻¹(f(x)) \[\frac{1}{\sqrt{1-\{f\left(x\right)\}^{2}}}\frac{d}{dx}f\left(x\right)\] |f(x)| < 1
cos⁻¹x \[-\frac{1}{\sqrt{1-x^{2}}}\] x| < 1
cos⁻¹(f(x)) \[-\frac{1}{\sqrt{1-\left\{f\left(x\right)\right\}^{2}}}\frac{d}{dx}f(x)\] |f(x)| < 1
tan⁻¹x \[\left(\frac{1}{1+x^{2}}\right)\] x ∈ R
tan⁻¹(f(x)) \[\frac{1}{1+\left\{f\left(x\right)\right\}^{2}}\frac{d}{dx}f(x)\] f(x) ∈ R
cot⁻¹x \[-\left(\frac{1}{1+x^{2}}\right)\] x ∈ R
cot⁻¹(f(x)) \[-\frac{1}{1+\{f(x)\}^{2}}\frac{d}{dx}f(x)\] f(x) ∈ R
sec⁻¹x \[\frac{1}{|x|\sqrt{x^{2}-1}}\] |x| > 1
sec⁻¹(f(x)) \[\frac{1}{|f(x)|\sqrt{\{f(x)\}^{2}-1}}\frac{d}{dx}f(x)\] |f(x)| > 1
cosec⁻¹x \[-\left(\frac{1}{|x|\sqrt{x^{2}-1}}\right)\]

|x| > 1

cosec⁻¹(f(x)) \[-\frac{1}{|f(x)|\sqrt{\{f(x)\}^{2}-1}}\frac{d}{dx}f(x)\] |f(x)| > 1
Formula: Left Hand and Right Hand Derivative

Left Derivative at x = c:

\[\lim_{h\to0^-}\frac{f(c+h)-f(c)}{h}\]

Right Derivative at x = c

\[\lim_{h\to0^+}\frac{f(c+h)-f(c)}{h}\]

Formula: Derivatives of Composite Functions

1. Chain Rule:

If u = g(x) and y = f(u), then

\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]

2. Composite Function Form:

If h(x) = f(g(x)), then

\[h^{\prime}(x)=f^{\prime}(g(x))\cdot g^{\prime}(x)\]

3. Power of a Function:

If y = [f(x)]n, then

\[\frac{dy}{dx}=n[f(x)]^{n-1}\cdot f^{\prime}(x)\]

4. Inverse Function Formula:

\[\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}},\quad\frac{dx}{dy}\neq0\]

\[\frac{dy}{dx}\cdot\frac{dx}{dy}=1\]

5. Derivative of Absolute Value Function:

For y=∣x∣,

\[\frac{d}{dx}(|x|)=\frac{x}{|x|},\quad x\neq0\]

6. Special Results:

\[\frac{d}{dx}(x)=1\]

\[\frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2},x\neq0\]

\[\frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}},\mathrm{~}x>0\]

\[\frac{d}{dx}(\sqrt{ax+b})=\frac{a}{2\sqrt{ax+b}}\]

Formula: Derivative at any Point

A function f is said to have a derivative at any point x if

\[f^{\prime}(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]

Formula: Implicit Differentiation

\[\frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}\]

\[\frac{d}{dx}(xy)=x\frac{dy}{dx}+y\]

Formula: Logarithmic Function
Function / Rule Derivative
log x \[\frac{1}{x}\]
\[\log_{a}x\] \[\frac{1}{x\log a}\]
\[\log_ax^n\] \[n\log_ax\]
log u \[\frac{1}{u}\cdot\frac{du}{dx}\]
\[log_a1\] 0
\[\log_aa\] 1
\[log_au\] \[\frac{1}{u\log a}\cdot\frac{du}{dx}\]
\[\log_a(xy)\] \[\log_ax+\log_ay\]
\[\log_a\left(\frac{x}{y}\right)\] \[\log_ax-\log_ay\]
\[\log_ax\] \[\frac{\log x}{\log a}\]
\[y=u^{v}\] \[u^v\frac{d}{dx}(v\log u)\]
Formula: Differentiation of a Determinant

For a 2×2 determinant:

\[F^{\prime}(x)=
\begin{vmatrix}
f_1^{\prime}(x) & f_2(x) \\
g_1^{\prime}(x) & g_2(x)
\end{vmatrix}+
\begin{vmatrix}
f_1(x) & f_2^{\prime}(x) \\
g_1(x) & g_2^{\prime}(x)
\end{vmatrix}\]

For a 3×3 determinant:

\[\mathrm{F^{\prime}}(x)=
\begin{vmatrix}
f_1^{\prime}(x) & f_2^{\prime}(x) & f_3^{\prime}(x) \\
g_1(x) & g_2(x) & g_3^{\prime}(x) \\
h_1(x) & h_2(x) & h_3(x)
\end{vmatrix}+
\begin{vmatrix}
f_1(x) & f_2(x) & f_3(x) \\
g_1^{\prime}(x) & g_2^{\prime}(x) & g_3^{\prime}(x) \\
h_1(x) & h_2(x) & h_3(x)
\end{vmatrix}+
\begin{vmatrix}
f_1(x) & f_2(x) & f_3(x) \\
g_1(x) & g_2(x) & g_3(x) \\
h_1^{\prime}(x) & h_2^{\prime}(x) & h_3^{\prime}(x)
\end{vmatrix}\]

Formula: Derivative of Quotient Function

Quotient Rule:

If \[y=\frac{u}{v}\] then \[\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\]

Reciprocal Rule:

\[\frac{d}{dx}{\left(\frac{1}{f(x)}\right)}=-\frac{f^{\prime}(x)}{[f(x)]^2}\]

Formula: Parametric Functions

First derivative:

If x = f(t), y = ϕ(t) then \[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]

Second derivative:

\[\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\]

Formula: Exponential Function
  • \[(a^x)^y=a^{xy}\]

  • \[a^{-x}=\frac{1}{a^x}\]

\[\frac{d}{dx}(e^x)=e^x\]

\[\frac{d}{dx}(a^x)=a^x\log a,\quad a>0,a\neq1\]

\[\frac{d}{dx}(e^{f(x)})=e^{f(x)}\cdot f^{\prime}(x)\]

\[\frac{d}{dx}(a^{f(x)})=a^{f(x)}\log a\cdot f^{\prime}(x)\]

Formula: Derivative of Inverse Trigonometric Functions

A. Trigonometric Functions

Function (y) \[\frac{dy}{dx}\]
sin x cos x
cos x -sin x
tan x sec2 x
cot x -cosec2 x
sec x sec x.tan x
cosec x  -cosec x cot x

B. Inverse Trigonometric Functions

Function Derivative
sin−1x \[\frac{1}{\sqrt{1-x^2}}\]
cos⁡−1x \[-\frac{1}{\sqrt{1-x^2}}\]
tan⁡−1x \[\frac{1}{1+x^2}\]
cot⁡−1x \[-\frac{1}{1+x^2}\]
sec⁡−1x \[\frac{1}{x\sqrt{x^2-1}}\]
cosec-1 \[-\frac{1}{x\sqrt{x^2-1}}\]
Formula: Derivative of Product of Function

(i) Product of two functions

If y = uv then, \[\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\]

(i) Product of three functions

If y = uvw then \[\frac{dy}{dx}=uv\frac{dw}{dx}+uw\frac{dv}{dx}+vw\frac{du}{dx}\]

Theorems and Laws [9]

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.)

If x = `e^(x/y)`, then prove that `dy/dx = (x - y)/(xlogx)`.

Given: x = `e^(x/y)`

Taking log on both the sides,

log x = `log e^(x/y)`

⇒ log x = `x/y log e`

⇒ log x = `x/y`  ...[∵ log e = 1]  ...(i)

Differentiating both sides w.r.t. x:

`d/dx log x = d/dx (x/y)`

⇒ `1/x = (y xx 1 - x xx dy/dx)/y^2`

⇒ `y^2 = xy - x^2 xx dy/dx`

⇒ `x^2 xx dy/dx = xy - y^2`

⇒ `dy/dx = (y(x - y))/x^2`

⇒ `dy/d = y/x xx ((x - y)/x)`

⇒ `dy/dx = 1/logx xx ((x - y)/x)   ...[∵ log x = x/y "from equation (i)"]`

`dy/dx = (x - y)/(xlogx)`

Hence proved.

If cos y = x cos (a + y), with cos a ≠ ± 1, prove that `dy/dx = cos^2(a+y)/(sin a)`.

cos y = x cos (a + y)

∴ x = `(cos y)/(cos (a + y))`

On differentiating with respect to y,

`cos (a + y) d/dy cos y - cos y d/dy`

`therefore dx/dy = (cos (a + y))/(cos^2 (a + y))`

`= (- sin y cos (a + y) + cos y sin (a + y))/(cos^2 (a + y))`

`= (sin (a + y) cos y - cos (a + y) sin y)/(cos^2 (a + y))`

`= (sin (a + y - y))/(cos^2 (a + y))`   ...[∵ sin (A − B) = sin A cos B − cos A sin B]

`= (sin a)/(cos^2  (a + y))`

`therefore dy/dx = (cos^2 (a + y))/(sin a)`

If y = 5 cos x – 3 sin x, prove that `(d^2y)/(dx^2) + y = 0`.

Given, y = 5 cos x – 3 sin x

Differentiating both sides with respect to x,

`dy/dx = 5 d/dx cos x - 3 d/dx sin x`

= 5 (−sin x) − 3 cos x

= −5 sin x − 3 cos x

Differentiating both sides again with respect to x,

`(d^2 y)/dx = - 5 d/dx sin x - 3 d/dx cos x`

= −5 cos x − 3 (−sin x)

= 3 sin x − 5 cos x

Hence, `(d^2 y)/dx^2 + y` = 0

(3 sin x − 5 cos x) + (5 cos x − 3 sin x) = 0 ...(On substituting the value of y)

Theorem: Lagrange's Mean Value Theorem

f a function f(x) is

  1. Continuous on [a,b]

  2. Differentiable on (a,b)

Then there exists at least one c ∈ (a,b) such that

\[f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\]

Theorem: Rolle’s Theorem

If a function f(x) is

  1. Continuous on [a,b]

  2. Differentiable on (a,b)

  3. f(a) = f(b)

Then there exists at least one c ∈ (a,b) such that f′(c) = 0

Key Points

Key Points: Algebra of Continuous Functions

If f & g are two continuous functions at x = a, then

αf is continuous at x = a ∀ α ∈ R
f + g is continuous at x = a
f − g is continuous at x = a
f·g is continuous at x = a
f/g is continuous at x = a, provided g(a) ≠ 0

Key Points: Derivative of Composite Functions

If y is a differentiable function of u and u is a differentiable function of x, then

\[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\cdot\frac{\mathrm{d}u}{\mathrm{d}x}\]

Key Points: Derivative of Inverse Functions

If y = f(x) is a differentiable function of x such that the inverse function x = f⁻¹(y) exists, then x is a differentiable function of y and

\[\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}\], where \[\frac{\mathrm{d}y}{\mathrm{d}x}\neq0\].

Key Points: Logarithmic Differentiation

If differentiation of an expression is done after taking the logarithm on both sides, then it is called logarithmic differentiation. Generally, we apply this method when the given expression is in one of the following forms:

  1. product of a number of functions,
  2. a quotient of functions,
  3. a function which is the power of another function, i.e., \[[f(x)]^{g(x)}\]
Key Points: When Function is Not Continuous

A function fails to be continuous at x = a if any one of the following occurs:

  1. f(a) is not defined

  2. \[\lim_{x\to a}f(x)\] does not exist

    • Either LHL or RHL does not exist

    • Or LHL ≠ RHL

  3. \[\lim_{x\to a}f(x)\] exists but \[\lim_{x\to a}f(x)\] ≠ f(a)

Key Points: Left Hand and Right Hand Continuity

(a) Left Hand Continuity at x = a

A function is left continuous at x = a if:

  1. f(a) exists

  2. \[\lim_{x\to a^-}f(x)\mathrm{~exists}\]

  3. \[\lim_{x\to a^-}f(x)=f(a)\]

(b) Right Hand Continuity at x = a

A function is right continuous at x = a if:

  1. f(a) exists

  2. \[\lim_{x\to a^+}f(x)\mathrm{~exists}\]

  3. \[\lim_{x\to a^+}f(x)=f(a)\]

c) Continuity at x = a

A function is continuous at x = a iff it is both left continuous and right continuous at x = a.

Key Points: Types of Discontinuity
Basis of Comparison Removable Discontinuity Non-Removable Discontinuity
Existence of \[\lim_{x\to a}f(x)\] Exists Does not exist
Left Hand Limit (LHL) Exists May not exist
Right Hand Limit (RHL) Exists May not exist
Relation between LHL & RHL LHL = RHL LHL ≠ RHL (or one/both do not exist)
Value of f(a) Not defined OR f(a) ≠ \[\lim_{x\to a}f(x)\] May or may not be defined
Continuity at ( x = a )  Discontinuous Discontinuous
Graphical interpretation Hole/gap in the graph Jump, break or vertical asymptote
Nature of discontinuity Temporary Permanent

Important Questions [47]

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