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

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

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. 

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)

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

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

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

(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 \[\frac{dy}{dx}=uv\frac{dw}{dx}+uw\frac{dv}{dx}+vw\frac{du}{dx}\]

Left Derivative at x = c:

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

Right Derivative at x = c

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

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:

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

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

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

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

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

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

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

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

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:

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

Quotient Rule:

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

Reciprocal Rule:

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

First derivative:

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

Second derivative:

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

• \[a^{0}=1\]​

• \[a^x\cdot a^y=a^{x+y}\]​

• \[(a^x)^y=a^{xy}\]

• \[a^{-x}=\frac{1}{a^x}\]

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

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

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

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

A. Trigonometric Functions

B. Inverse Trigonometric Functions

f(x) = xⁿ

`lim_(x->n)` f(x) = `lim_(x -> n)` xⁿ = nⁿ

f(n) = nⁿ

`lim_(x -> n)` f(x) = f(n)

f is continuous at x = n, where n is a positive integer.

⇒ f is continuous at n ∈ N.

The given function is:

f(x) = 5x – 3

f(0) = 5(0) – 3 = –3

`lim_(x → 0)` f(x) = 5(0) – 3 = –3

`lim_(x → 0)` f(x) = f(0)

Hence, the function is continuous at x = 0.

f(–3) = 5(–3) – 3

= –15 – 3

= –18

⇒ `lim_(x → -3)` f(x) = 5(–3) – 3

= –15 – 3

= –18

⇒ `lim_(x → -3)` f(x) = f(–3)

Hence, the function is continuous at x = –3.

f(5) = 5(5) – 3

= 25 – 3

= 22

⇒ `lim_(x → 5)` f(x)

= 5(5) – 3

= 25 – 3

= –22

⇒ `lim_(x -> 5)` f(x) = f(5)

Hence, the function is continuous at x = 5.

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.

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.

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)|`

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.)

`x sqrt(1 + y) + y sqrt(1 + x) = 0`

∴ `xsqrt(1 + y) = - y sqrt(1 + x) = 0`

On squaring both sides,

x² (1 + y) = y² (1 + x)

⇒ x² + x²y = y² + y²x

⇒ x² – y² – y²x + x²y = 0

⇒ (x – y)(x + y) + xy(x – y) = 0

⇒ (x – y)[x + y + xy] = 0

x – y = 0 ⇒ x ≠ y

x + y (1 + x) = 0

∴ y = `-x/(1 - x)`

∴ `dy/dx = ((1 + x)(1) - x * 1)/(1 + x)^2`

= `-(1 + x - x)/(1 + x)^2`

= `-1/(1 + x)^2`

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.

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)`

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)

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

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

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​

(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.

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)