CBSE Class 12 Mathematics Continuity and Differentiability Notes Set 03

These notes cover Chapter 5 Continuity and Differentiability of Class 12 Mathematics (CBSE) with important definitions, formulas, solved examples with step marks, Do Yourself practice questions and their answers. They take you from continuity at a point to differentiability, implicit, logarithmic, parametric and second order derivatives.

Continuity: Important Notes, Formulas and Solved Examples

1. Continuous Functions

A function for which a small change in the independent variable causes only a small change and not a sudden jump in the dependent variable is called continuous function. Naively, we may say that a function is continuous at a fixed point if we can draw the graph of the function around that point without lifting the pen from the plane of the paper.

2. Continuity of a Function at a Point

A function f(x) is said to be continuous at x = a, if \( \lim_{x \to a} f(x) = f(a) \). Symbolically f is continuous at x = a if \( \lim_{h \to 0} f(a-h) = \lim_{h \to 0} f(a+h) = f(a) \), h > 0

3. Reasons of Discontinuity

(a) Limit does not exist
i.e. \( \lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x) \)
(b) f(x) is not defined at x = a
(c) \( \lim_{x \to a} f(x) \neq f(a) \)
Geometrically, the graph of the function will exhibit a break at x = a, if the function is discontinuous at x = a. The graph as shown is discontinuous at x = 1, 2 and 3.

4. Continuous Functions

A function is said to be continuous function if it is continuous at every point in its domain.

Illustration 1. If \( f(x) = \begin{cases} \frac{1-\cos 4x}{8x^2}, & x \neq 0 \\ k, & x = 0 \end{cases} \), is continuous at x = 0, then find the value of k. [Exemplar] [4M]
Answer:
It is given that the function f is continuous at x = 0. Therefore, \( \lim_{x \to 0} f(x) = f(0) \) [1]
\( \Rightarrow \lim_{x \to 0} \frac{1-\cos 4x}{8x^2} = k \quad \Rightarrow \lim_{x \to 0} \frac{2\sin^2 2x}{8x^2} = k \) [1½]
\( \Rightarrow \lim_{x \to 0} \left(\frac{\sin 2x}{2x}\right)^2 = k \quad \Rightarrow k = 1 \) [1½]
Thus, f is continuous at x = 0 if k = 1.

Illustration 2. If \( f(x) = \frac{\sqrt{2}\cos x - 1}{\cot x - 1}, x \neq \frac{\pi}{4} \), find the value of \( f\left(\frac{\pi}{4}\right) \) so that f(x) becomes continuous at \( x = \frac{\pi}{4} \). [Exemplar] [4M]
Answer:
Given \( f(x) = \frac{\sqrt{2}\cos x - 1}{\cot x - 1}, x \neq \frac{\pi}{4} \)
Therefore, \( \lim_{x \to \frac{\pi}{4}} f(x) = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\cos x - 1}{\cot x - 1} \)
\( = \lim_{x \to \frac{\pi}{4}} \frac{(\sqrt{2}\cos x - 1)\sin x}{\cos x - \sin x} \)
\( = \lim_{x \to \frac{\pi}{4}} \frac{(\sqrt{2}\cos x - 1)}{(\sqrt{2}\cos x + 1)} \cdot \frac{(\sqrt{2}\cos x + 1)}{(\cos x - \sin x)} \cdot \frac{(\cos x + \sin x)}{(\cos x + \sin x)} \cdot \sin x \) [1]
\( = \lim_{x \to \frac{\pi}{4}} \frac{2\cos^2 x - 1}{\cos^2 x - \sin^2 x} \cdot \frac{\cos x + \sin x}{\sqrt{2}\cos x + 1} \cdot (\sin x) \)
\( = \lim_{x \to \frac{\pi}{4}} \frac{\cos 2x}{\cos 2x} \cdot \left(\frac{\cos x + \sin x}{\sqrt{2}\cos x + 1}\right) \cdot (\sin x) \) [1]
\( = \lim_{x \to \frac{\pi}{4}} \frac{(\cos x + \sin x)}{\sqrt{2}\cos x + 1} \sin x = \frac{\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)}{\sqrt{2} \cdot \frac{1}{\sqrt{2}} + 1} = \frac{1}{2} \) [1]
Thus, \( \lim_{x \to \frac{\pi}{4}} f(x) = \frac{1}{2} \)
Hence for f to be continuous at \( x = \frac{\pi}{4}, f\left(\frac{\pi}{4}\right) = \frac{1}{2} \) [1]

Illustration 3. Let \( f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}} - 4}, & x > 0 \end{cases} \)
For what value of a, is 'f' continuous at x = 0? [Exemplar] [4M]
Answer:
Now, f(0) = a; \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1-\cos 4x}{x^2} = \lim_{x \to 0^-} \frac{2\sin^2 2x}{x^2} = \lim_{x \to 0^-} 8\left(\frac{\sin 2x}{2x}\right)^2 = 8 \) [1]
and \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}} - 4} = \lim_{x \to 0^+} \frac{\sqrt{x}\left(\sqrt{16+\sqrt{x}} + 4\right)}{\left(\sqrt{16+\sqrt{x}} - 4\right)\left(\sqrt{16+\sqrt{x}} + 4\right)} \) [1]
\( = \lim_{x \to 0^+} \frac{\sqrt{x}\left(\sqrt{16+\sqrt{x}} + 4\right)}{16 + \sqrt{x} - 16} = \lim_{x \to 0^+} \left(\sqrt{16+\sqrt{x}} + 4\right) = \sqrt{16} + 4 = 8 \) [1]
Hence, \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = 8 = f(0) \)
∴ f is continuous at x = 0 if a = 8. [1]

Illustration 4. Find the value of a and b such that the function f defined by \( f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & \text{if } x < 4 \\ a + b, & \text{if } x = 4 \\ \frac{x-4}{|x-4|} + b, & \text{if } x > 4 \end{cases} \) is a continuous function at x = 4. [Exemplar] [4M]
Answer:
\( \lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} \frac{x-4}{|x-4|} + b \)
\( = \lim_{x \to 4^+} \frac{x-4}{x-4} + b = 1 + b \) {∵ x > 4, |x − 4| = (x − 4)} [1]
and \( \lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} \frac{x-4}{|x-4|} + a \)
\( = \lim_{x \to 4^-} \frac{x-4}{-(x-4)} + a = -1 + a \) {∵ x < 4, |x − 4| = −(x − 4)} [1]
Now f(4) = a + b [1]
Thus \( \lim_{x \to 4^+} f(x) = \lim_{x \to 4^-} f(x) = f(4) \)
1 + b = −1 + a = a + b
a − b = 2 and b = −1
So a = 1, b = −1 [1]

Do Yourself 1

(i) Find all the points of discontinuity of the function: \( f(x) = \begin{cases} |x| + 3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \ge 3 \end{cases} \) [4M]
(ii) If function \( f(x) = \begin{cases} \frac{1-\cos kx}{x \sin x}, & \text{if } x \neq 0 \\ \frac{1}{2}, & \text{if } x = 0 \end{cases} \) is continuous at x = 0, then find the value of k. [Exemplar] [4M]
(iii) Show that the function f given by \( f(x) = \begin{cases} \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} \) is discontinuous at x = 0. [4M]

5. Some Important Points

Suppose f(x) and g(x) be two real functions continuous at a real number a, then
(i) f(x) ± g(x) is continuous at x = a.
(ii) f(x) . g(x) is continuous at x = a.
(iii) \( \left(\frac{f(x)}{g(x)}\right) \) is continuous at x = a, (provided g(a) ≠ 0).
(iv) Suppose f and g are real valued functions such that (f o g) is defined at x = a. If g is continuous at x = a and if f is continuous at g(a) then (f o g) is continuous at x = a.

Illustration 5. Show that the function f defined by f(x) = |1 − x + |x||, where x is any real number, is a continuous function. [4M]
Answer:
Define g by g(x) = 1 − x + |x| and h by h(x) = |x| for all real x. Then
(h o g)(x) = h(g(x))
= h(1 − x + |x|)
= |1 − x + |x|| = f(x) [2]
We have seen that h is a continuous function. Hence g being a sum of a polynomial function and the modulus function is continuous. Hence f being a composite of two continuous functions is continuous. [2]

Illustration 6. If \( f(x) = \frac{x+1}{x-1} \) and \( g(x) = \frac{1}{x-2} \), then discuss the continuity of f(x), g(x) and fog(x) in R. [4M]
Answer:
f(x) is a rational function, it must be continuous in its domain and f is not defined at x = 1.
∴ f is discontinuous at x = 1 [1]
g(x) is also a rational function. It must be continuous in its domain and g is not defined at x = 2.
∴ g is discontinuous at x = 2 [1]
Now fog(x) will be discontinuous at x = 2 (point of discontinuity of g(x))
Consider g(x) = 1 (when g(x) = point of discontinuity of f(x))
\( \frac{1}{x-2} = 1 \quad \Rightarrow \quad x = 3 \)
∴ fog(x) is discontinuous at x = 2 & x = 3. [2]

Differentiability and Differentiation: Important Notes, Formulas and Solved Examples

1. Meaning of Derivative

The instantaneous rate of change of a function with respect to the independent variable is called derivative. Let ‘f’ be a given function of one variable and let Δx denote a number (positive or negative) to be added to the number x. Let Δf denote the corresponding change of ‘f’ then Δf = f(x + Δx) − f(x)
\( \Rightarrow \frac{\Delta f}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x} \)
If Δf/Δx approaches a limit as Δx approaches zero, this limit is the derivative of ‘f’ at the point x. The derivative of a function ‘f’ is a function; this function is denoted by symbols such as
\( f'(x), \frac{df}{dx}, \frac{d}{dx}\big(f(x)\big) \) or if y = f(x) by \( \frac{dy}{dx} \) or y'.
\( \Rightarrow \frac{df}{dx} = \lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \)
The process of finding derivative of a function is called differentiation. We also use the phrase differentiate f(x) with respect to x to mean find f'(x).

2. Differentiability: Existence of Derivative at x = a

(a) Right hand derivative:
The right hand derivative of f(x) at x = a denoted by \( f'(a^+) \) is defined as:
\( f'(a^+) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \), provided the limit exists & is finite. (h > 0)
(b) Left hand derivative:
The left hand derivative of f(x) at x = a denoted by \( f'(a^-) \) is defined as:
\( f'(a^-) = \lim_{h \to 0} \frac{f(a-h) - f(a)}{-h} \), provided the limit exists & is finite. (h > 0)
Hence f(x) is said to be derivable or differentiable at x = a if RHD = LHD = finite quantity and it is denoted by f'(a); where \( f'(a) = f'(a^-) = f'(a^+) \) & it is called derivative or differential coefficient of f(x) at x = a.

Note:
(i) Right hand and left hand derivative at x = a is also denoted by Rf'(a) and Lf'(a) respectively.
(ii) The common value of Rf'(a) and Lf'(a) is denoted by f'(a).

Illustration 7. Consider the function \( f(x) = \begin{cases} x[x], & \text{if } 0 \le x < 2 \\ (x-1)x, & \text{if } 2 \le x < 3 \end{cases} \).
Discuss derivability of f at x = 2. [Exemplar] [4M]
Answer:
Here, f(2) = (2 − 1) 2 = 1 × 2 = 2,
LHD = \( \lim_{h \to 0} \frac{f(a-h) - f(a)}{-h} = \frac{(2-h)[2-h] - 2}{-h} = \frac{(2-h) - 2}{-h} = 1 \) [1½]
RHD = \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0} \frac{(2+h-1)(2+h) - 2}{h} \)
\( = \lim_{h \to 0} \frac{h^2 + 3h}{h} = \lim_{h \to 0} \frac{h(h+3)}{h} = \lim_{h \to 0} (h + 3) = 3 \) [1½]
Thus, LHD ≠ RHD ⇒ f is not derivable at x = 2. [1]

Illustration 8. Prove that the function f given by f(x) = |x − 1|, x ∈ R is not differentiable at x = 1. [4M]
Answer:
Given f(x) = |x − 1|
We redefine the function as
\( f(x) = \begin{cases} -x + 1, & \text{if } x < 1 \\ x - 1, & \text{if } x \ge 1 \end{cases} \)
Now, at x = 1
LHD = \( \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} \)
\( = \lim_{h \to 0} \frac{-(1-h) + 1 - (1-1)}{-h} \)
\( = \lim_{h \to 0} \frac{-1 + h + 1 - 0}{-h} = \lim_{h \to 0} \frac{h}{-h} = -1 \) [1½]
RHD = \( \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \)
\( = \lim_{h \to 0} \frac{1 + h - 1 - 0}{h} = \lim_{h \to 0} \frac{h}{h} = 1 \) [1½]
LHD ≠ RHD
So, f(x) is not differentiable at x = 1. [1]

3. Theorem

If a function f(x) is derivable at x = a, then f(x) is continuous at x = a, i.e. every differentiable function is continuous.
Proof: \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \) exists.
Also \( f(a+h) - f(a) = \frac{f(a+h) - f(a)}{h} \cdot h \quad [h \neq 0] \)
\( \therefore \lim_{h \to 0} [f(a+h) - f(a)] = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \cdot h = f'(a) \cdot 0 = 0 \)
\( \Rightarrow \lim_{h \to 0} [f(a+h) - f(a)] = 0 \Rightarrow \lim_{h \to 0} f(a+h) = f(a) \Rightarrow \) f(x) is continuous at x = a.

Note:
(i) Converse of above theorem is NOT true.
(ii) All polynomial, trigonometric, logarithmic and exponential functions are continuous and differentiable in their domains.
(iii) If f(x) & g(x) are differentiable at x = a then the functions f(x) + g(x), f(x) − g(x), f(x) . g(x) will also be differentiable at x = a & if g(a) ≠ 0 then the function f(x)/g(x) will also be differentiable at x = a.

4. Derivability Over an Interval

(a) f(x) is said to be derivable over an open interval (a, b) if it is derivable at each & every point of the open interval (a, b).
(b) f(x) is said to be derivable over the closed interval [a, b] if:
(i) f(x) is derivable in (a, b) &
(ii) for the points a and b, \( f'(a^+) \) & \( f'(b^-) \) exist.

Illustration 9. Consider the function \( f(x) = \begin{cases} 1 + x, & \text{if } x \le 2 \\ 5 - x, & \text{if } x > 2 \end{cases} \).
Check the continuity and differentiability of f(x) at x = 2. [Exemplar] [4M]
Answer:
Here, f(2) = 1 + 2 = 3.
\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (1 + x) = 1 + 2 = 3 \) and \( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (5 - x) = 5 - 2 = 3. \)
Thus, \( f(2) = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 3 \)
⇒ f is continuous at x = 2. [2]
Also, \( Lf'(2) = \lim_{h \to 0} \frac{f(2-h) - f(2)}{-h} = \lim_{h \to 0} \frac{1 + (2-h) - 3}{-h} = \lim_{h \to 0} \frac{-h}{-h} = 1 \)
and \( Rf'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0} \frac{5 - (2+h) - 3}{h} = \lim_{h \to 0} \frac{-h}{h} = -1 \)
⇒ Lf'(2) ≠ Rf'(2) ⇒ f is not derivable at x = 2. [2]

Illustration 10. Examine the differentiability of the function f defined by \( f(x) = \begin{cases} 2x + 3, & \text{if } -3 \le x < -2 \\ x + 1, & \text{if } -2 \le x < 0 \\ x + 2, & \text{if } 0 \le x \le 1 \end{cases} \) [Exemplar] [4M]
Answer:
The only doubtful points for differentiability of f(x) are x = −2 and x = 0. Differentiability at x = −2.
Now \( Lf'(-2) = \lim_{h \to 0} \frac{f(-2-h) - f(-2)}{-h} \)
\( = \lim_{h \to 0} \frac{2(-2-h) + 3 - (-2+1)}{-h} = \lim_{h \to 0} \frac{-2h}{-h} = \lim_{h \to 0} 2 = 2 \)
and \( Rf'(-2) = \lim_{h \to 0} \frac{f(-2+h) - f(-2)}{h} \)
\( = \lim_{h \to 0} \frac{-2 + h + 1 - (-2+1)}{h} \)
\( = \lim_{h \to 0} \frac{h - 1 - (-1)}{h} = \lim_{h \to 0} \frac{h}{h} = 1 \)
Thus Rf'(−2) ≠ Lf'(−2). Therefore f is not differentiable at x = −2. [2]
Similarly, for differentiability at x = 0, we have
\( Lf'(0) = \lim_{h \to 0} \frac{f(0-h) - f(0)}{-h} \)
\( = \lim_{h \to 0} \frac{0 - h + 1 - (0 + 2)}{-h} \)
\( = \lim_{h \to 0} \frac{-h - 1}{-h} = \lim_{h \to 0} \left(1 + \frac{1}{h}\right) \)
which does not exist. Hence f is not differentiable at x = 0. [2]

Do Yourself 2

(i) Examine the differentiability of f, where f is defined by \( f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} \) at x = 0. [Exemplar] [4M]
(ii) Find f'(0) for the function \( f(x) = \begin{cases} \frac{1 - \cos x}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases} \) [4M]

5. Differentiation

Fundamental Theorems:
If f and g are derivable functions of x, then,
(a) \( \frac{d}{dx}\big(f(x) \pm g(x)\big) = f'(x) \pm g'(x) \)
(b) \( \frac{d}{dx}\big(cf(x)\big) = cf'(x) \), where c is any constant
(c) \( \frac{d}{dx}\big(f(x)\, g(x)\big) = f(x)\, g'(x) + g(x)\, f'(x) \) known as “PRODUCT RULE”
(d) \( \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)\, f'(x) - f(x)\, g'(x)}{\big(g(x)\big)^2} \), where g(x) ≠ 0 known as “QUOTIENT RULE”
(e) If y = f(u) & u = g(x), then \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \) known as “CHAIN RULE”
Note: In general if y = f(u), then \( \frac{dy}{dx} = f'(u) \cdot \frac{du}{dx} \).

6. Derivative of Standard Functions

Illustration 11. Differentiate the function \( \frac{\sin(ax + b)}{\cos(cx + d)} \) with respect to x. [4M]
Answer:
Let \( y = \frac{\sin(ax + b)}{\cos(cx + d)} \)
\( \Rightarrow \frac{dy}{dx} = \frac{a\cos(cx + d)\cos(ax + b) + c\sin(ax + b)\sin(cx + d)}{\cos^2(cx + d)} \) [1]
\( \Rightarrow \frac{dy}{dx} = \frac{a\cos(cx + d)\cos(ax + b)}{\cos^2(cx + d)} + \frac{c\sin(ax + b)\sin(cx + d)}{\cos^2(cx + d)} \) [1]
\( \Rightarrow \frac{dy}{dx} = a\cos(ax + b)\sec(cx + d) + c\sin(ax + b)\tan(cx + d)\sec(cx + d) \) [2]

Illustration 12. Differentiate the function \( 2\sqrt{\cot(x^2)} \) with respect to x. [Exemplar] [4M]
Answer:
Let \( y = 2\sqrt{\cot x^2} \)
\( \Rightarrow \frac{dy}{dx} = 2 \times \frac{1}{2\sqrt{\cot x^2}} \times \frac{d}{dx}(\cot x^2) \) [1]
\( \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{\cot x^2}} \times -\text{cosec}^2 x^2 \times \frac{d}{dx}(x^2) \)
\( \Rightarrow \frac{dy}{dx} = \frac{-\sqrt{\sin x^2}}{\sqrt{\cos x^2}} \times \frac{1}{\sin^2 x^2} \times 2x \) [1]
\( \Rightarrow \frac{dy}{dx} = \frac{-2x}{\sqrt{\sin x^2 \cos x^2}} \times \frac{1}{\sin x^2} \times \frac{\sqrt{2}}{\sqrt{2}} \) [1]
\( \Rightarrow \frac{dy}{dx} = \frac{-2\sqrt{2}x}{\sqrt{2\sin x^2 \cos x^2}\, \sin x^2} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-2\sqrt{2}x}{\sin x^2 \sqrt{\sin 2x^2}} \) [1]

Illustration 13. If \( y = \log\left[x + \sqrt{x^2 + a^2}\right] \), then find \( \frac{dy}{dx} \). [Exemplar] [4M]
Answer:
We have \( y = \log\left[x + \sqrt{x^2 + a^2}\right] \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \times \frac{d}{dx}\left[x + \sqrt{x^2 + a^2}\right] \) [1]
\( \Rightarrow \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \times \left[1 + \frac{1}{2\sqrt{x^2 + a^2}} \times \frac{d}{dx}(x^2 + a^2)\right] \) [1]
\( \Rightarrow \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \times \left[1 + \frac{1}{2\sqrt{x^2 + a^2}} \times 2x\right] \) [½]
\( \Rightarrow \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \times \left[\frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}}\right] \) [1]
\( \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{x^2 + a^2}} \) [½]

Illustration 14. If \( y = \log\sqrt{\frac{1 + \sin x}{1 - \sin x}} \), then find \( \frac{dy}{dx} \). [4M]
Answer:
\( y = \log\sqrt{\frac{1 + \sin x}{1 - \sin x}} \)
\( y = \log\sqrt{\frac{1 + \sin x}{1 - \sin x} \times \frac{1 + \sin x}{1 + \sin x}} \) [1]
\( \Rightarrow y = \log\sqrt{\frac{(1 + \sin x)^2}{1 - \sin^2 x}} \)
\( \Rightarrow y = \log\sqrt{\frac{(1 + \sin x)^2}{\cos^2 x}} \) [½]
\( \Rightarrow y = \log\left(\frac{1 + \sin x}{\cos x}\right) \)
\( \Rightarrow y = \log(\sec x + \tan x) \) [1]
Diff. w.r.t. x
\( \Rightarrow \frac{dy}{dx} = \frac{1}{\sec x + \tan x} \times \frac{d}{dx}(\sec x + \tan x) \) [½]
\( \Rightarrow \frac{dy}{dx} = \frac{1}{\sec x + \tan x} \times (\sec x \tan x + \sec^2 x) \)
\( \Rightarrow \frac{dy}{dx} = \frac{\sec x(\tan x + \sec x)}{(\sec x + \tan x)} \)
\( \Rightarrow \frac{dy}{dx} = \sec x \) [1]

Do Yourself 3

Differentiate the following functions w.r.t. x [4M each]
(i) \( \frac{2^x \cot x}{\sqrt{x}} \)
(ii) \( \frac{e^x + e^{-x}}{e^x - e^{-x}} \)
(iii) \( y = \log\left(\frac{\sin x}{1 + \cos x}\right) \)
(iv) \( \log\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right) \)

7. Differentiation of Implicit Function

Let the function be φ(x, y) = 0. Then to find dy/dx in the case of implicit functions, we differentiate each term w.r.t. x regarding y as a function of x & then collect terms in dy/dx together on one side to finally find dy/dx.

Illustration 15. If y = x sin y, prove that \( x\frac{dy}{dx} = \frac{y}{1 - x\cos y} \) [4M]
Answer:
Given y = x sin y, differentiating both sides w.r.t. x.
\( \frac{dy}{dx} = x\cos y \frac{dy}{dx} + \sin y \frac{d}{dx}(x) \quad \Rightarrow \frac{dy}{dx} - x\cos y \frac{dy}{dx} = \sin y \times 1 \) [1½]
\( \Rightarrow \frac{dy}{dx} = \frac{\sin y}{1 - x\cos y} \), multiplying both sides by x [1]
\( x\frac{dy}{dx} = \frac{x\sin y}{1 - x\cos y} = \frac{y}{1 - x\cos y} \quad (\because x\sin y = y) \) [1½]

Illustration 16. If \( \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \), then show that \( \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}} \) [Exemplar] [4M]
Answer:
Given \( \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \)
Putting x = sin α ⇒ \( \alpha = \sin^{-1} x \), and y = sin β ⇒ \( \beta = \sin^{-1} y \), we get
\( \sqrt{1 - \sin^2 \alpha} + \sqrt{1 - \sin^2 \beta} = a(\sin \alpha - \sin \beta) \)
⇒ cos α + cos β = a(sin α − sin β) [1]
\( \Rightarrow 2\cos\left(\frac{\alpha + \beta}{2}\right)\cos\left(\frac{\alpha - \beta}{2}\right) = a \cdot 2\cos\left(\frac{\alpha + \beta}{2}\right)\sin\left(\frac{\alpha - \beta}{2}\right) \) [1]
\( \Rightarrow \cot\left(\frac{\alpha - \beta}{2}\right) = a \Rightarrow \frac{\alpha - \beta}{2} = \cot^{-1} a \Rightarrow \alpha - \beta = 2\cot^{-1} a \)
\( \Rightarrow \sin^{-1} x - \sin^{-1} y = 2\cot^{-1} a \) [1]
Differentiating both sides w.r.t. x, we get
\( \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}} \) [1]

Illustration 17. If \( e^x + e^y = e^{x+y} \), prove that \( \frac{dy}{dx} = -e^{y-x} \). [Exemplar] [4M]
Answer:
Given that \( e^x + e^y = e^{x+y} \). Diff. w.r.t. x we get
\( e^x + e^y \frac{dy}{dx} = e^{x+y}\left(1 + \frac{dy}{dx}\right) \) [1]
\( \Rightarrow (e^y - e^{x+y})\frac{dy}{dx} = e^{x+y} - e^x \) [1]
\( \Rightarrow \frac{dy}{dx} = \frac{e^{x+y} - e^x}{e^y - e^{x+y}} = \frac{e^x + e^y - e^x}{e^y - e^x - e^y} = -e^{y-x} \quad [\because e^x + e^y = e^{x+y}] \) [2]

8. Derivatives of Inverse Trigonometric Functions

Illustration 18. Find \( \frac{dy}{dx} \) when \( y = \sin^{-1}\left(\frac{\sqrt{x} - 1}{\sqrt{x} + 1}\right) + \sec^{-1}\left(\frac{\sqrt{x} + 1}{\sqrt{x} - 1}\right), x > 1 \). [Exemplar] [4M]
Answer:
Given \( y = \sin^{-1}\left(\frac{\sqrt{x} - 1}{\sqrt{x} + 1}\right) + \sec^{-1}\left(\frac{\sqrt{x} + 1}{\sqrt{x} - 1}\right) \)
\( = \sin^{-1}\left(\frac{\sqrt{x} - 1}{\sqrt{x} + 1}\right) + \cos^{-1}\left(\frac{\sqrt{x} - 1}{\sqrt{x} + 1}\right) \quad \left(\because \sec^{-1} x = \cos^{-1}\frac{1}{x}, |x| \ge 1\right) \) [1½]
\( = \frac{\pi}{2} \quad \left(\because \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}, |x| \le 1\right) \) [1½]
\( \Rightarrow \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) = 0 \) [1]

Illustration 19. If \( y = \sin^{-1}\left\{\frac{\sqrt{1 + x} - \sqrt{1 - x}}{2}\right\} \), then find \( \frac{dy}{dx} \). [Exemplar] [4M]
Answer:
\( y = \sin^{-1}\left\{\frac{\sqrt{1 + x} - \sqrt{1 - x}}{2}\right\} \)
Put \( x = \cos 2\theta;\ \theta = \frac{1}{2}\cos^{-1} x \) [1]
\( \Rightarrow y = \sin^{-1}\left\{\frac{\sqrt{1 + \cos 2\theta} - \sqrt{1 - \cos 2\theta}}{2}\right\} \)
\( \Rightarrow y = \sin^{-1}\left\{\frac{\sqrt{2\cos^2 \theta} - \sqrt{2\sin^2 \theta}}{2}\right\} \) [1]
\( \Rightarrow y = \sin^{-1}\left\{\frac{1}{\sqrt{2}}\cos\theta - \frac{1}{\sqrt{2}}\sin\theta\right\} \)
\( \Rightarrow y = \sin^{-1}\left\{\sin\frac{\pi}{4}\cos\theta - \cos\frac{\pi}{4}\sin\theta\right\} \)
\( \Rightarrow y = \sin^{-1}\left\{\sin\left(\frac{\pi}{4} - \theta\right)\right\} \) [1]
\( \Rightarrow y = \frac{\pi}{4} - \theta \)
\( \Rightarrow y = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x \)
Diff. w.r.t. x.
\( \Rightarrow \frac{dy}{dx} = -\frac{1}{2} \times \frac{-1}{\sqrt{1 - x^2}} \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{1 - x^2}} \) [1]

Illustration 20. \( y = \tan^{-1}\left[\frac{\cos x}{1 + \sin x}\right] \), find \( \frac{dy}{dx} \). [4M]
Answer:
\( y = \tan^{-1}\left[\frac{\cos x}{1 + \sin x}\right] \)
\( \Rightarrow y = \tan^{-1}\left[\frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2}\right] \) [1]
\( \Rightarrow y = \tan^{-1}\left[\frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}\right] \)
\( \Rightarrow y = \tan^{-1}\left[\frac{1 - \tan\frac{x}{2}}{1 + \tan\frac{x}{2}}\right] \) [1]
\( \Rightarrow y = \tan^{-1}\left[\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right] \) [1]
\( \Rightarrow y = \frac{\pi}{4} - \frac{x}{2} \)
Diff. w.r.t. x
\( \frac{dy}{dx} = -\frac{1}{2} \) [1]

Illustration 21. \( y = \sin^{-1}\left(\frac{2x}{1 + x^2}\right) \), |x| ≤ 1, find \( \frac{dy}{dx} \). [4M]
Answer:
Let \( y = \sin^{-1}\left(\frac{2x}{1 + x^2}\right) \), put x = tan θ ⇒ \( \theta = \tan^{-1} x \) [1]
then \( y = \sin^{-1}\left(\frac{2\tan\theta}{1 + \tan^2 \theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1} x \) [2]
differentiating w.r.t. x, we get
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1 + x^2} = \frac{2}{1 + x^2} \) [1]
Note: The above solution is valid only for x ∈ [−1, 1]. It may be noted that the given function is defined for all x ∈ R. To get a more general solution, differentiate directly using chain rule as explained below:
\( \frac{d}{dx}\left\{\sin^{-1}\left(\frac{2x}{1 + x^2}\right)\right\} = \frac{1}{\sqrt{1 - \left(\frac{2x}{1 + x^2}\right)^2}} \cdot \frac{d}{dx}\left(\frac{2x}{1 + x^2}\right) \)
\( = \frac{1 + x^2}{\sqrt{(1 - x^2)^2}} \cdot \frac{2(1 + x^2) - 2x \cdot 2x}{(1 + x^2)^2} = \frac{2(1 - x^2)}{|1 - x^2|\,(1 + x^2)}, \ |x| \neq 1 \)
So \( \frac{dy}{dx} = \frac{2}{1 + x^2} \) for |x| < 1 and \( \frac{dy}{dx} = \frac{-2}{1 + x^2} \) for |x| > 1.

Do Yourself 4

Find \( \frac{dy}{dx} \) when [4M each]
(i) \( y = \cos(2\sin^{-1} x) + \sin(2\sin^{-1} x) \)
(ii) \( y = \sin^{-1}\sqrt{\frac{1 + x^2}{2}} \)
(iii) \( y = \cos^{-1}\left(\frac{x - x^{-1}}{x + x^{-1}}\right) \)

9. Logarithmic Differentiation

To find the derivative of:
(a) A function which is the product or quotient of a number of functions or
(b) A function of the form \( [f(x)]^{g(x)} \) where f & g are both derivable, it is convenient to take the logarithm of the function first & then differentiate.
Consider \( y = f(x) = [u(x)]^{v(x)} \)
By taking logarithm (to base e) the above may be rewritten as log y = v(x) log[u(x)]
Using chain rule we may differentiate this to get
\( \frac{1}{y} \cdot \frac{dy}{dx} = v(x) \cdot \frac{1}{u(x)} \cdot u'(x) + v'(x) \cdot \log[u(x)] \)
\( \Rightarrow \frac{dy}{dx} = y\left[\frac{v(x)}{u(x)} \cdot u'(x) + v'(x) \cdot \log[u(x)]\right] \).

Illustration 22. \( y = \frac{x\sqrt{x^2 + 4}}{(3x + 2)^{2/3}}, x > 0 \). Find \( \frac{dy}{dx} \). [4M]
Answer:
Let \( y = \frac{x\sqrt{x^2 + 4}}{(3x + 2)^{2/3}} \), taking (natural) logarithm on the two sides, we have
\( \log y = \log\left\{\frac{x\sqrt{x^2 + 4}}{(3x + 2)^{2/3}}\right\} = \log x + \frac{1}{2}\log(x^2 + 4) - \frac{2}{3}\log(3x + 2) \) [1]
Differentiating both sides w.r.t. x, we obtain
\( \frac{1}{y}\frac{dy}{dx} = \frac{1}{x} + \frac{1}{2} \cdot \frac{1}{x^2 + 4}(2x + 0) - \frac{2}{3} \cdot \frac{1}{3x + 2}(3 \times 1 + 0) \) [1½]
\( \Rightarrow \frac{dy}{dx} = y\left\{\frac{1}{x} + \frac{x}{x^2 + 4} - \frac{2}{3x + 2}\right\} \) or \( \frac{dy}{dx} = \frac{x\sqrt{x^2 + 4}}{(3x + 2)^{2/3}}\left\{\frac{1}{x} + \frac{x}{x^2 + 4} - \frac{2}{3x + 2}\right\} \) [1½]

Illustration 23. Differentiate the function \( x^{\sin x} + (\sin x)^{\cos x} \) w.r.t. x. [4M]
Answer:
Let \( y = x^{\sin x} + (\sin x)^{\cos x} \)
and \( u = x^{\sin x} \), \( v = (\sin x)^{\cos x} \)
Now, y = u + v
and \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ........(i) [1]
We have, \( u = x^{\sin x} \)
Taking log on both sides, we have
log u = sin x log x
Differentiating both sides w.r.t. x, we have
\( \frac{1}{u}\frac{du}{dx} = \sin x \cdot \frac{1}{x} + \log x \cdot \cos x \)
\( \frac{du}{dx} = u\left(\frac{\sin x}{x} + \log x \cdot \cos x\right) \)
\( = x^{\sin x}\left[\frac{\sin x}{x} + \log x \cdot \cos x\right] \) [1]
Again \( v = (\sin x)^{\cos x} \)
Taking log on both sides; we have
log v = cos x log sin x
Differentiating both sides w.r.t. x, we have
\( \frac{1}{v}\frac{dv}{dx} = \cos x \cdot \frac{1}{\sin x} \cdot \cos x + \log \sin x\,(-\sin x) \)
\( = \cos x \cdot \cot x + \log \sin x \cdot (-\sin x) \)
\( = \cos x \cdot \cot x - \sin x \cdot \log \sin x \)
\( \therefore \frac{dv}{dx} = v[\cos x \cdot \cot x - \sin x \cdot \log \sin x] \)
\( \frac{dv}{dx} = (\sin x)^{\cos x}[\cos x \cdot \cot x - \sin x \log \sin x] \) [1]
Putting these values in (i), we have
\( \frac{dy}{dx} = x^{\sin x}\left[\frac{\sin x}{x} + \log x \cdot \cos x\right] + (\sin x)^{\cos x}[\cos x \cot x - \sin x \log \sin x] \) [1]

Do Yourself 5

[4M each]
(i) If \( x^y = e^{x-y} \), prove that \( \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} \)
(ii) Differentiate \( \cos(x^x) \) w.r.t. x.
(iii) Differentiate \( (\sin x)^x + (\cos x)^{\tan x} \) w.r.t. x.

10. Parametric Differentiation

If y = f(θ) & x = g(θ) where θ is a parameter, then \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \).

Illustration 24. If \( x = e^{\cos 2t} \) and \( y = e^{\sin 2t} \), prove that \( \frac{dy}{dx} = \frac{-y\log x}{x\log y} \). [Exemplar] [4M]
Answer:
\( x = e^{\cos 2t} \)
log x = cos 2t .........(1)
\( \Rightarrow \frac{1}{x} \cdot \frac{dx}{dt} = -2\sin 2t \Rightarrow \frac{dx}{dt} = -2x\sin 2t \) [1½]
and \( y = e^{\sin 2t} \)
log y = sin 2t .........(2)
\( \Rightarrow \frac{1}{y} \cdot \frac{dy}{dt} = 2\cos 2t \)
\( \Rightarrow \frac{dy}{dt} = 2y\cos 2t \) [1½]
Now \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = -\frac{2y\cos 2t}{2x\sin 2t} = -\frac{y\log x}{x\log y} \) (from equation (1) and (2)). [1]

Do Yourself 6

[4M each]
(i) If \( x = a\sec^3 \theta \) and \( y = a\tan^3 \theta \), then find the value of \( \frac{d^2y}{dx^2} \) at \( \theta = \frac{\pi}{4} \).
(ii) If \( x = a\sqrt{\cos 2t} \cdot \cos t \) and \( y = a\sqrt{\cos 2t} \cdot \sin t \), then find the value of \( \frac{dy}{dx} \) at \( t = \frac{\pi}{6} \).

11. Second Order Derivative

Let y = f(x). Then
\( \frac{dy}{dx} = f'(x) \) ...(i)
If f'(x) is differentiable, we may differentiate (i) again w.r.t. x. Then, the left hand side becomes \( \frac{d}{dx}\left(\frac{dy}{dx}\right) \) which is called the second order derivative of y w.r.t. x and is denoted by \( \frac{d^2y}{dx^2} \).
The second order derivative of f(x) is denoted by f''(x). It is also denoted by \( D^2y \) or y'' or \( y_2 \) if y = f(x). Similarly, higher order derivatives may be defined.

Illustration 25. (i) Find the second derivative of log x + cos x.
(ii) If y = tan x + sec x, prove that \( \frac{d^2y}{dx^2} = \frac{\cos x}{(1 - \sin x)^2} \). [4M]
Answer:
(i) Let y = log x + cos x, differentiating w.r.t. x, we get
\( \frac{dy}{dx} = \frac{1}{x} - \sin x \), again differentiating w.r.t. x, we get
\( \frac{d^2y}{dx^2} = (-1)x^{-2} + (-\cos x) = -\frac{1}{x^2} - \cos x \) [2]
(ii) Given y = sec x + tan x, differentiating w.r.t. x, we get
\( \frac{dy}{dx} = \sec x\tan x + \sec^2 x = \frac{1}{\cos x}\left(\frac{\sin x}{\cos x}\right) + \frac{1}{\cos^2 x} = \frac{\sin x + 1}{\cos^2 x} = \frac{1 + \sin x}{1 - \sin^2 x} \)
\( = \frac{1 + \sin x}{(1 - \sin x)(1 + \sin x)} = \frac{1}{1 - \sin x} \)
Again differentiating w.r.t. x, we have
\( \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{1}{1 - \sin x}\right) \)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{d}{dx}\{(1 - \sin x)^{-1}\} = (-1)(1 - \sin x)^{-2}(0 - \cos x) = \frac{\cos x}{(1 - \sin x)^2} \). [2]

Illustration 26. If \( y = e^{a\cos^{-1} x}, -1 \le x \le 1 \), show that \( (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2 y = 0 \) [4M]
Answer:
We have \( y = e^{a\cos^{-1} x} \) .............(1)
Diff. w.r.t. x
\( \frac{dy}{dx} = e^{a\cos^{-1} x} \times \frac{-a}{\sqrt{1 - x^2}} \quad \Rightarrow \frac{dy}{dx} = \frac{-ay}{\sqrt{1 - x^2}} \) from equation (1) [1]
\( \Rightarrow \sqrt{1 - x^2} \cdot \frac{dy}{dx} = -ay \quad \Rightarrow (1 - x^2)\left(\frac{dy}{dx}\right)^2 = a^2y^2 \) [1]
Diff. w.r.t. x.
\( \Rightarrow (1 - x^2) \times 2\left(\frac{dy}{dx}\right) \times \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 \times (-2x) = 2a^2y \cdot \frac{dy}{dx} \) [1]
\( \Rightarrow \frac{2dy}{dx}\left[(1 - x^2)\frac{d^2y}{dx^2} - x \cdot \frac{dy}{dx}\right] = 2a^2y \cdot \frac{dy}{dx} \quad \Rightarrow (1 - x^2)\frac{d^2y}{dx^2} - x \cdot \frac{dy}{dx} - a^2y = 0 \) [1]

Do Yourself 7

(i) If y = a cos mx + b sin mx, prove that \( y_2 + m^2 y = 0 \)
(ii) If \( y = (\sin^{-1} x)^2 \), prove that \( (1 - x^2)y_2 - xy_1 = 2 \).
(iii) If \( y = \frac{\sin^{-1} x}{x} \), then prove that \( x(1 - x^2)y_2 + (2 - 3x^2)y_1 - xy = 0 \)

Answers for Do Yourself

Do Yourself 1: (i) Discontinuous at x = 3 (ii) k = ±1
Do Yourself 2: (i) Differentiable (ii) 1/2
Do Yourself 3: (i) \( -\frac{2^{x-1}\cot x}{x\sqrt{x}} - \frac{2^x \text{cosec}^2 x}{\sqrt{x}} + \frac{2^x \cot x \ln 2}{\sqrt{x}} \) (ii) \( \frac{-4}{(e^x - e^{-x})^2} \) (iii) cosec x (iv) cosec x
Do Yourself 4: (i) \( -4x + \frac{2(1 - 2x^2)}{\sqrt{1 - x^2}} \) (ii) \( \frac{x}{\sqrt{1 - x^4}} \) (iii) \( \frac{-2}{1 + x^2} \)
Do Yourself 5: (ii) \( -x^x(1 + \log x)\sin(x^x) \) (iii) \( (\sin x)^x(x\cot x + \log \sin x) + (\cos x)^{\tan x}(\sec^2 x \log \cos x - \tan^2 x) \)
Do Yourself 6: (i) \( \frac{1}{12a} \) (ii) 0