ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.9

Interactive ISC Mathematics Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.9 engage students in active learning and exploration.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.9

Differentiate the following functions (1 to 7) w.r.t. x :

Question 1.
(i) (x + 3)² (x + 4)³ (x + 5)⁴ (NCERT)
(ii) cos x . cos 2x . cos 3x (NCERT)
Solution:
(i) Let y = (x + 3)² (x + 4)³ (x + 5)⁴
Taking logarithm on both sides of eqn. (1) ; we get
log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)
[∵ log a^b = b log a
and log ab = log a + log b]
Diff. bothsides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\)
∴ \(\frac{d y}{d x}\) = (x + 3)² (x + 4)³ (x + 5)⁴ \(\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]\) [using (1)]

(ii) Let y = cos x . cos 2x . cos 3x …………….(1)
Taking logarithm on both sides of eqn. (1) ; we get
log y = log cos x + log cos 2x + log cos 3x ;
Diff. bothsides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=\frac{1}{\cos x}(-\sin x)+\frac{1}{\cos 2 x}(-2 \sin 2 x)+\frac{1}{\cos 3 x}(-3 \sin 3 x)\)
∴ \(\frac{d y}{d x}\) = cos x cos 2x cos 3x . [- tan x – 2 tan 2x – 3 tan 3x] [using (1)]

Question 2.
(i) e^x cos³ x sin² x
(ii) \(\frac{e^{x^2} \tan ^{-1} x}{\sqrt{1+x^2}}\).
Solution:
(i) Let y = e^x cos³ x sin² x
Taking logarithm on both sides, we have ;
log y = x log e + 3 log cos x + 2 log sin x
Diff. bothsides w.r.t. x, we get
∴ \(\frac{1}{y} \frac{d y}{d x}=1+\frac{3}{\cos x}(-\sin x)+\frac{2}{\sin x}(\cos x)\)
Thus \(\frac{d y}{d x}\) = e^x cos³ x sin² x [1 – 3 tan x + 2 cot x] [using (1)]

(ii) Let y = \(\frac{e^{x^2} \tan ^{-1} x}{\sqrt{1+x^2}}\)
Taking logarithm on both sides, we get ;
log y = log e^x2 + log tan^-1 x – \(\frac{1}{2}\) log (1 + x²)
⇒ log y = x² + log tan^{-1</sup x – \(\frac{1}{2}\) log (1 + x2)
Diff. bothsides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=2 x+\frac{1}{\tan ^{-1} x} \frac{1}{1+x^2}-\frac{1}{2} \times \frac{1}{1+x^2} \times 2 x\)
⇒ \(\frac{d y}{d x}=\frac{e^{x^2} \tan ^{-1} x}{\sqrt{1+x^2}}\left[2 x+\frac{1}{\left(1+x^2\right) \tan ^{-1} x}-\frac{x}{1+x^2}\right]\)}

Question 2 (old).
(i) \(\frac{x \sqrt{x^2+1}}{(x+1)^{2 / 3}}\), x > 0
(ii) \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\) (NCERT)
Solution:
(i) Let y = \(\frac{x \sqrt{x^2+1}}{(x+1)^{2 / 3}}\), x > 0
Taking natural logarithm on both sides ; we get
log y = log x + \(\frac{1}{2}\) log (x² + 1) – \(\frac{2}{3}\) log (x + 1)
[[∵ log a^b = b log a ;
log \(\frac{a}{b}\) = log a – log b
and log ab = log a + log b]
Diff. bothsides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=\frac{1}{x}+\frac{1}{2} \frac{1}{x^2+1} \times 2 x-\frac{2}{3(x+1)}\)
∴ \(\frac{d y}{d x}=y\left[\frac{1}{x}+\frac{x}{x^2+1}-\frac{2}{3(x+1)}\right]\) ; x ≠ 0
= \(\frac{x \sqrt{x^2+1}}{(x+1)^{2 / 3}}\left[\frac{1}{x}+\frac{x}{x^2+1}-\frac{2}{3(x+1)}\right]\) ; x ≠ 0

(ii) Let y = \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\)
Taking natural logarithm on both sides ; we get
log y = \(\frac{1}{2}\) [log (x – 1) + log (x – 2) – log (x – 3) – log (x – 4) – log (x – 5)]
[using properties of logarithm]
Diff. both sides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=\frac{1}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]\)
⇒ \(\frac{d y}{d x}=\frac{y}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]\)
= \(\frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]\).

Question 3.
(i) (sin x)^{cos x}, 0 < x < π (NCERT Exampler)
(ii) (sin x)^{sin x}, 0 < x< π (NCERT)
Solution:
(i) Let y = (sin x)^{cos x} …………..(1)
Taking natural logarithm on both sides ; we have
log y = cos x. log (sin x)
Differentiate both sides w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}\) = cos x \(\frac{1}{\sin x}\) cos x + log (sin x) (- sin x)
⇒ \(\frac{d y}{d x}=\frac{(\sin x)^{\cos x}}{\sin x}\) [cos² x – sin² x log (sin x)] [using (1)\

(ii) Let y = (sin x)^{sin x} ; 0 < x< π
Taking natural logarithm on both sides ; we have
log y = sin x log sin x ;
Diff. both sides w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}\) = sin x \(\frac{1}{\sin x}\) cos x + (log sin x) cos x
⇒ \(\frac{d y}{d x}\) = y cos x [1 + log (sin x)]
⇒ \(\frac{d y}{d x}\) = (sin x)^{sin x} cos x [1 + log sin x].

Question 4.
(i) x^{sin x}, x > 0 (NCERT)
(ii) (2x + 3)^{x – 5}, x > – \(\frac{3}{2}\)
Solution:
(i) Let y = x^{sin x}, x > 0
Taking logarithm on both sides we get ;
log y = sin x . log x ;
Differentiate w.r.t. x
∴ \(\frac{1}{y} \frac{d y}{d x}\) = sin x . \(\frac{1}{x}\) + log x cos x
⇒ \(\frac{d y}{d x}\) = x^{sin x} [\(\frac{sin x}{x}\) + log x . cos x] [using (1)]

(ii) Let y = (2x + 3)^{x – 5}, x > – \(\frac{3}{2}\)
Taking logarithm on both sides we get ;
log y = (x – 5) log (2x + 3)
Diff. both sides w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}\) = \(\frac{x-5}{2 x+3}\) . 2 + log (2x + 3) . 1
[using product rule]
⇒ \(\frac{d y}{d x}\) = (2x + 3)^{x – 5} [\(\frac{2(x-5)}{2 x+3}\) + log (2x + 3)]

Question 5.
(i) (log x)^{cos x}, x > 1 (NCERT)
(ii) (log x)^{log x}, x > 1 (NCERT)
Solution:
(i) Let y = (log x)^{cos x}
Taking logarithm on both sides ; we have
log y = cos x . log (log x)
Differentiate both sides w.r.t. x, we have
∴ \(\frac{1}{y} \frac{d y}{d x}\) = cos x . \(\frac{1}{\log x} \cdot \frac{1}{x}\) + log (log x) (- sin x)
⇒ \(\frac{d y}{d x}\) = (log x)^{cos x} [\(\frac{\cos x}{x \log x}\) – sin x . log (log x)] [using (1)]

(ii) Let y = (log x)^{log x}
Taking logarithm on both sides ; we have
log y = log x . log (log x)
Differentiate both sides w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}\) = log x . \(\frac{1}{\log x} \cdot \frac{1}{x}\) + log (log x) . \(\frac{1}{x}\)
⇒ \(\frac{1}{y} \frac{d y}{d x}\) = \(\frac{1}{x}\) [1 + log (log x)]
∴ \(\frac{d y}{d x}\) = \(\frac{(\log x)^{\log x}}{x}\) [1 + log (log x)]

Question 6.
(i) If y = x^y, prove that x \(\frac{d y}{d x}\) = \(\frac{y^2}{1-y \log x}\).
(ii) If x = e^x/y, prove that \(\frac{d y}{d x}=\frac{x-y}{x \log x}\). (NCERT Exampler)
(iii) If x^y = e^x-y, prove that \(\frac{d y}{d x}=\frac{\log x}{(\log x e)^2}\).
(iv) If x¹⁶ y⁹ = (x² + y)¹⁷, prove that \(\frac{d y}{d x}=\frac{2 y}{x}\).
Solution:
(i) Given y = x^y ………(1)
Taking logarithm on both sides ; we get ;
log y = y log x ;
Differentiate both sides w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}=\frac{y}{x}+\log x \frac{d y}{d x}\)
⇒ \(\left(\frac{1}{y}-\log x\right) \frac{d y}{d x}=\frac{y}{x}\)
⇒ \(\left(\frac{1-y \log x}{y}\right) \frac{d y}{d x}=\frac{y}{x}\)
⇒ \(\frac{d y}{d x}=\frac{y^2}{x(1-y \log x)}\)

(ii) Given x = e^x/y ;
Taking logarithm on both sides ; we get ;
log x = log e^x/y
= \(\frac{x}{y}\) . 1
⇒ y log x = x
On differentiating both sides w.r.t. x ; we get
\(\frac{d}{d x}\) (y log x) = \(\frac{d}{d x}\) (x)
⇒ y \(\frac{d}{d x}\) log x + log x \(\frac{d y}{d x}\) = 1
⇒ \(\frac{y}{x}\) + log x \(\frac{d y}{d x}\) = 1
⇒ log x \(\frac{d y}{d x}\) = 1 – \(\frac{y}{x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{x-y}{x \log x}\)

(iii) Given, x^y = e^x-y ;
Taking logarithm on both sides ; we get ;
y log x = (x – y) log e = x – y …………(1)
⇒ y (1 + log x) = x
⇒ y = \(\frac{x}{1+\log x}\) ……….(2)
Diff. eqn. (2) both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{(1+\log x) \cdot 1-x\left(0+\frac{1}{x}\right)}{(1+\log x)^2}\)
= \(\frac{\log x}{(1+\log x)^2}\)
= \(\frac{(x-y)}{y[\log e+\log x]^2}\) [using eqn. (1)]
= \(\frac{x-y}{y(\log e x)^2}\)
[∵ log a + log b = log ab]

(iv) Given, x¹⁶ y⁹ = (x² + y)¹⁷
Taking logarithm on both sides ; we get ;
log (x¹⁶ y⁹) = log (x² + 17)¹⁷
⇒ log x¹⁶ + log y⁹ = 17 log (x² + y)
⇒ 16 log x + 9 log y = 17 log (x² + y)
Diff. both sides w.r.t. x ; we get

Question 7.
Find the derivative of x^x + a^x + x^a + a^a for some fixed a > 0, x > 0. (NCERT)
Solution:
Let y = x^x + a^x + x^a + a^a, a > 0, x > 0
Diff. both sides w.r.t. x ; we get
\(\frac{d y}{d x}=\frac{d}{d x} x^x+\frac{d}{d x} a^x+\frac{d}{d x} x^a+\frac{d}{d x} a^a\)
= \(\frac{d}{d x}\) e log x^x + a^x log a + ax^a-1 + 0
[∵ a^a be a constant
∴ \(\frac{d}{d x}\) (a^a) = 0]
= e^{x log x} [x × \(\frac{1}{x}\) + log x . 1] + a^x log a + ax^a-1
= x^x [1 + log x] + a^x log a + ax^a-1

Question 7 (old).
(i) x^{sin x + cos x}
(ii) (x² sin x)^1/x
Solution:
(i) Let y = x^{sin x + cos x}
Taking logarithm on both sides ; we have
log y = (sin x + cos x) log x
Differentiate both sides w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}\) = (sin x + cos x) \(\frac{1}{x}\) + log x (cos x – sin x)
\(\frac{d y}{d x}\) = x^{sin x + cos x} [\(\frac{\sin x+\cos x}{x}\) + (cos x – sin x) log x]

(ii) Let y = (x² sin x)^1/x
Taking logarithm on both sides ; we have
log y = \(\frac{1}{x}\) log (x² sin x)
Differentiate both sides w.r.t. x, we have

Question 8.
Differentiate the following functions w.r.t. x:
(i) x^{log x} + (log x)^x
(ii) (sin x)^{cos x} + x^{sin x}
(iii) x^{cos x} + (cos x)^x
(iv) (x)^{cos x} + (sin x)^{tan x}
(v) (sin 2x)^x + sin^-1 \(\sqrt{3x}\)
(vi) e^{sin x} + (tan x)^x
(vii) x^x – 2^{sin x} (NCERT)
(viii) x^{x cos x} + \(\frac{x^2+1}{x^2-1}\)
Solution:
(i) Let y = x^{log x} + (log x)^x
= u + v
where u = x^{log x} ; v = (log x)^x
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………….(1)
Now u = x^{log x}
so that log u = (log x)²
Diff. both sides w.r.t. x ; we get
\(\frac{1}{u} \frac{d u}{d x}=\frac{2 \log x}{x}\)
⇒ \(\frac{d u}{d x}=x^{\log x}\left(\frac{2 \log x}{x}\right)\) …………(2)
Also v = (log x)^x
so that log v = x log (log x)
Differentiate both sides w.r.t. x, we have
\(\frac{1}{v} \frac{d v}{d x}=x \cdot \frac{1}{\log x} \cdot \frac{1}{x}\) + log (log x) . 1
⇒ \(\frac{d v}{d x}\) = (log x)^x [\(\frac{1}{\log x}\) + log (log x)] ………….(3)
Putting eqn. (2) and (3) in eqn. (1) ; we get
∴ \(\frac{d y}{d x}=x^{\log x}\left(\frac{2 \log x}{x}\right)+(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right]\)

(ii) Let y = (sin x)^{cos x} + (x)^{sin x} = u + v ;
where u = sin x^{cos x} ;
v = (x)^{sin x}
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
Now u = (sin x)^{cos x} so that
log u = cos x log (sin x) ;
Diff. both sides w.r.t. x ; we get
\(\frac{1}{u} \frac{d u}{d x}\) = cos x . cot x + log (sin x) (- sin x)
∴ \(\frac{d u}{d x}\) = (sin x)^{cos x} [cos x cot x – sin x log (sin x)] ………..(2)
Also v = (x)^{sin x}
so that log v = sin x . log x
Diff. both sides w.r.t. x ; we get
\(\frac{1}{v} \frac{d v}{d x}\) = \(\frac{sin x}{x}\) + cos x log x
∴ \(\frac{d v}{d x}\) = x^{sin x} [\(\frac{sin x}{x}\) + cos x log x]………….(3)
putting the values of eqn. (2) and (3) in eqn. (1), we have
\(\frac{d y}{d x}\) = (sin x)^{cos x} [cos x cot x – sin x log (sin x)] + x^{sin x} [\(\frac{sin x}{x}\) + cos x log x]

(iii) Let y = x^{cos x} + (cos x)^x
= u + v ;
where u = x^{cos x} ;
v = (cos x)^x
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
Now u = x^{cos x}
so that log u = cos x log x ;
Diff. both sides w.r.t. x ; we have
\(\frac{1}{u} \frac{d u}{d x}\) = cos x . \(\frac{1}{x}\) + log x (- sin x)
⇒ \(\frac{d u}{d x}\) = x^{cos x} [\(\frac{cos x}{x}\) – sin x log x] ……..(2)
also, v = (cos x)^x so that
log v = x log cos x ;
Diff. both sides w.r.t. x ;
\(\frac{1}{v} \frac{d u}{d x}\) = – x tan x + log (cos x)
⇒ \(\frac{d v}{d x}\) = (cos x)^x [- x tan x + log (cos x)] …………..(3)
putting eqn. (2) and eqn. (3) in eqn. (1) ; we get
\(\frac{d y}{d x}\) = x^{cos x} [\(\frac{cos x}{x}\) – sin x log x] + (cos x)^x [- x tan x + log (cos x)]

(iv) Let y = x^{cos x} + (sin x)^{tan x}
= u + v
where u = x^{cos x} ; v = (sin x)^{tan x}
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
Now u = x^{cos x} so that log x = cos x log x
Differentiate both sides w.r.t. x ; we have
∴ \(\frac{1}{u} \frac{d u}{d x}\) = \(\frac{cos x}{x}\) – sin x log x
∴ \(\frac{d u}{d x}\) = x^{cos x} [\(\frac{cos x}{x}\) – sin x log x] ……….(2)
Also v = (sin x)^{tan x} so that
log v = tan x log sin x ;
Differentiate w.r.t. x
\(\frac{1}{v} \frac{d v}{d x}\) = tan x . cot x + log (sin x) sec² x
\(\frac{d v}{d x}\) = (sin x)^{tan x} [1 + sec² log sin x]
putting the values of eqn’s (2) and (3) in eqn. (1) ; we have
∴ \(\frac{d y}{d x}\) = x^{cos x} [\(\frac{cos x}{x}\) – sin x log x] + (sin x)^{tan x} [1 + sec² log sin x]

(v) Let y = (sin 2x)^x + sin^-1 \(\sqrt{3x}\)
= u + v ………….(1)
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(2)
Given u = (sin 2x)^x ;
Taking logarithm on both sides ; we have
log u = x log (sin 2x)^x = x log (sin 2x) ;
Diff. both sides w.r.t. x
∴ \(\frac{1}{u} \frac{d u}{d x}\) = log (sin 2x) + \(\frac{x}{sin 2x}\) . 2 cos 2x
⇒ \(\frac{d u}{d x}\) = (sin 2x)^x [log sin 2x + 2x cot 2x]
and v = sin^-1 \(\sqrt{3x}\) ………..(3)
∴ \(\frac{d v}{d x}\) = \(\frac{1}{\sqrt{1-(\sqrt{3} x)^2}} \frac{d}{d x}\) √3 √x
= \(\frac{1}{\sqrt{1-3 x}} \frac{\sqrt{3}}{2} \frac{1}{\sqrt{x}}\) …………..(4)
putting eqn. (3) and eqn. (4) in eqn. (2) ; we have
∴ \(\frac{d y}{d x}\) = (sin 2x)^x [log sin 2x + 2x cot 2x] + \(\frac{\sqrt{3}}{2 \sqrt{x-3 x^2}}\)

(vi) Let y = e^{sin x} + (tan x)^x
= u + v
where u = e^{sin x}
and v = (tan x)^x
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
Now u = e^{sin x}
⇒ \(\frac{d u}{d x}\) = e^{sin x} cos x …………..(2)
Also v = (tan x)^x so that log v = x log (tan x)
Differentiate both sides w.r.t. x ; we have
\(\frac{1}{v} \frac{d v}{d x}\) = [log (tan x) + x \(\frac{\sec ^2 x}{\tan x}\)]
⇒ \(\frac{d v}{d x}\) = (tan x)^x [log (tan x) + x sec² x cot x] ……………(3)
Putting (2) and (3) in eqn. (1) ; we get
∴ \(\frac{d y}{d x}\) = e^{sin x} cos x + (tan x)^x [log (tan x) + x sec x cosec x]

(vii) Let y = x^x – 2^{sin x} ;
Diff. both sides w.r.t. x, we get
∴ \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x^x) – 2^{sin x} log 2 . cos x ………..(1)
Let u = x^x so that
log u = x log x
∴ \(\frac{d u}{d x}\) = x^x (1 + log x) ………..(2)
Putting eqn. (2) in (1) ; we get
∴ \(\frac{d y}{d x}\) = x^x (1 + log x) – cos x . 2^{sin x} log 2.

(viii) Let y = x^{x cos x} + \(\frac{x^2+1}{x^2-1}\)
⇒ y = e^{log xx cos x} + \(\frac{x^2+1}{x^2-1}\)
⇒ y = e^{(x cos x) log x} + \(\frac{x^2+1}{x^2-1}\)

Question 8 (old).
(v) (sin x)^x + sin^-1 √x
Solution:
Let y = (sin x)^x + sin^-1 √x
= u + v ;
where u = (sin x)^x
and v = sin^-1 √x
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
Now u = (sin x)^x so that log u = x . log (sin x)
Differentiate both sides w.r.t. x, we get
\(\frac{1}{u} \frac{d u}{d x}\) = log (sin x) . 1 + x cot x
⇒ \(\frac{d u}{d x}\) = (sin x)^x [log (sin x) + x cot x] ………….(2)
Now v = sin^-1 √x
⇒ \(\frac{d v}{d x}\) = \(\frac{1}{\sqrt{1-(\sqrt{x})^2}} \frac{1}{2 \sqrt{x}}\)
= \(\frac{1}{2 \sqrt{x-x^2}}\) ………..(3)
putting the values of eqn’s (2) and (3) in eqn. (1) ; we have
∴ \(\frac{d y}{d x}\) = (sin x)^x [log (sin x) + x cot x] + \(\frac{1}{2 \sqrt{x-x^2}}\)

Question 9.
(i) If y = (log x)^{cos x} + \(\frac{x^2+1}{x^2-1}\), find \(\frac{d y}{d x}\).
(ii) If y = e^{x2 cos x} + (cos x)^x, find \(\frac{d y}{d x}\).
Solution:
Let y = (log x)^{cos x} + \(\frac{x^2+1}{x^2-1}\)
= u + v,
where u = (log x)^{cos x}
and v = \(\frac{x^2+1}{x^2-1}\)
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
Now u = (log x)^{cos x}
so that log u = cos x (log log x) ;
Differentiate both sides w.r.t. x, we have
\(\frac{1}{u} \frac{d u}{d x}\) = cos x . \(\frac{1}{x \log x}\) + log (log x) (- sin x)
⇒ \(\frac{d u}{d x}\) = (log x)^{cos x} [\(\frac{\cos x}{x \log x}\) – sin x log (log x)] ………..(2)
Also v = \(\frac{x^2+1}{x^2-1}\) ;
Diff. both sides w.r.t. x, we get
\(\frac{d v}{d x}\) = \(\frac{\left(x^2-1\right) 2 x-\left(x^2+1\right) 2 x}{\left(x^2-1\right)^2}\)
= \(\frac{-4 x}{\left(x^2-1\right)^2}\) …………..(3)
Putting eqn. (2) and (3) in eqn. (1) ; we have
\(\frac{d y}{d x}\) = (log x)^{cos x} [\(\frac{\cos x}{x \log x}\) – sin x log (log x)] – \(\frac{4 x}{\left(x^2-1\right)^2}\)

(ii) Given y = e^{x2 cos x} + (cos x)^x
⇒ y = u + v
⇒ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………..(1)
where u = e^{x2 cos x}
diff. both sides w.r.t. x
\(\frac{d u}{d x}\) = e^{x2 cos x} \(\frac{d}{d x}\) (x² cos x)
= e^{x2 cos x} [x² (- sin x) + cos x . 2x] ……………..(2)
and v = (cos x)^x ;
taking logarithm on both sides
log v = x log cos x
∴ \(\frac{1}{v} \frac{d v}{d x}\) = log cos x + \(\frac{x}{cos x}\) (- sin x)
= log cos x – x tan x
⇒ \(\frac{d v}{d x}\) = (cos x)^x [log cos x – x tan x] ……….(3)
putting equation (2) and (3) in equation (1) ; we have
\(\frac{d y}{d x}\) = e^{x2 cos x} [- x² sin x + 2x cos x] + (cos x)^x [log cos x – x tan x]

Question 10.
Find \(\frac{d y}{d x}\) when
(i) x^y y^x = a^b
(ii) x^y + y^x = log a^b
(iii) x^x + y^x = 1
(iv) (sin x)^y = x + y
(v) xy = e^{x – y} (NCERT)
(vi) (cos x)^y = (cos y)^x
Solution:
(i) Given, x^y y^x = a^b ;
Taking logarithm both sides ; we have
log (x^y . y^x) = log a^b
⇒ log x^y + log y^x = 0
⇒ y log x + x log y = 0
Diff. both sides w.r.t. x ; we get
y \(\frac{d}{d x}\) log x + log x \(\frac{d y}{d x}\) + \(\frac{x}{y} \frac{d y}{d x}\) + log y \(\frac{d}{d x}\) (x) = 0
⇒ \(\frac{y}{x}\) + log x \(\frac{d y}{d x}\) + \(\frac{x}{y} \frac{d y}{d x}\) + log y . 1 = 0
⇒ (log x + \(\frac{y}{x}\)) \(\frac{d y}{d x}\) = – (log y + \(\frac{y}{x}\))
⇒ \(\frac{d y}{d x}=-\frac{y(x \log y+y)}{x(y \log x+x)}\)

(ii) Given x^y + y^x = log a^b
⇒ u + v = log a^b ……………(1)
where u = x^y
and v = y^x
Diff. eqn. (1) w.r.t. x ; we have
\(\frac{d u}{d x}\) + \(\frac{d v}{d x}\) = 0 …………(2)
Since u = x^y ;
Taking logarithm on both sides, we have
log u = y log x ;
Diff. both sides w.r.t. x, we have
\(\frac{1}{u} \frac{d u}{d x}\) = \(\frac{y}{x}\) + log x \(\frac{d y}{d x}\)
⇒ \(\frac{d u}{d x}\) = x^y [\(\frac{y}{x}\) + log x \(\frac{d y}{d x}\)]
= yx^y-1 + x^y log x \(\frac{d y}{d x}\)
Also v = y^x ;
Taking logarithm on both sides ; we have
log v = log y^x
= x log y
Diff. both sides w.r.t. x ; we have
\(\frac{1}{v} \frac{d v}{d x}\) = \(\frac{x}{y} \frac{d y}{d x}\) + log y . 1
⇒ \(\frac{d v}{d x}\) = y^x [\(\frac{x}{y} \frac{d y}{d x}\) + log y]
= xy^x-1 \(\frac{d y}{d x}\) + y^x log y
Using eqn. (3) and eqn. (4) in eqn. (2) ; we have
yx^y-1 + x^y log x \(\frac{d y}{d x}\) + xy^x-1 \(\frac{d y}{d x}\) + y^x log y = 0
⇒ (x^y log x + x y^{x – 1}) \(\frac{d y}{d x}\) = – (y x^x-1 + y^x log y)
⇒ \(\frac{d y}{d x}\) = – \(\frac{y\left(x^{y-1}+y^{x-1} \log y\right)}{x\left(x^{y-1} \log x+y^{x-1}\right)}\).

(iii) Given x^y + y^x = 1
⇒ u + v = 1
where u = x^x
and v = y^x
∴ \(\frac{d u}{d x}\) + \(\frac{d v}{d x}\) = 0 ………..(1)
Now u = x^x
so that lo g u = x log x
Differentiate both sides w.r.t. x, we get
\(\frac{1}{u} \frac{d u}{d x}\) = x . \(\frac{1}{x}\) + log x
⇒ \(\frac{d u}{d x}\) = x^x (1 + log x)
Also v = y^x
so that log v = x log y
Differentiate both sides w.r.t. x, we get
\(\frac{1}{v} \frac{d v}{d x}\) = \(\frac{x}{y} \frac{d y}{d x}\) + log y
⇒ \(\frac{d v}{d x}\) = y^x [\(\frac{x}{y} \frac{d y}{d x}\) + log y] ………….(30
putting eqn. (2) and (3) in eqn. (1) ; we get
x^x (1 + log x) + xy^x-1 \(\frac{d y}{d x}\) + y^x log y = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{-\left[y^x \log y+x^x(1+\log x)\right]}{x y^{x-1}}\).

(iv) Given (sin x)^y = x + y ;
Taking logarithm on both sides ; we have
log (sin x)^y = log (x + y)
⇒ y log sin x = log (x + y)
Diff. both sides w.r.t. x ; we get
\(\frac{d}{d x}\) y log sin x = \(\frac{d}{d x}\) log (x + y)
⇒ y \(\frac{d}{d x}\) log sin x + log sin x \(\frac{d y}{d x}\) = \(\frac{1}{x+y}\) \(\frac{d}{d x}\) (x + y)
⇒ \(\frac{y}{\sin x} \frac{d}{d x}\) (sin x) + log sin x \(\frac{d y}{d x}\) = \(\frac{1}{x+y}\left[1+\frac{d y}{d x}\right]\)
⇒ [y cot x + log sin x \(\frac{d y}{d x}\)] (x + y) = 1 + \(\frac{d y}{d x}\)
⇒ [(x + y) log sin x – 1] \(\frac{d y}{d x}\) = 1 – y (x + y) cot x
∴ \(\frac{d y}{d x}\) = \(\frac{1-y(x+y) \cot x}{(x+y) \log \sin x-1}\).

(v) Given xy = e^x-y ……….(1),
Taking logarithm on both sides, we get
log x + log y = (x – y) log e ;
Differentiate boht sides w.r.t. x, we have
\(\frac{1}{x}+\frac{1}{y} \frac{d y}{d x}=1-\frac{d y}{d x}\)
⇒ \(\left(\frac{1}{y}+1\right) \frac{d y}{d x}=1-\frac{1}{x}\)
⇒ \(\frac{d y}{d x}=\frac{(x-1) y}{(y+1) x}\)

(vi) Given (cos x)^y = (cos y)^x ;
Taking logarithm on both sides, we have
y log cos x = x log cos y
Diff. both sides w.r.t. x ; we get
⇒ \(\frac{d}{d x}\) [y log cos x] = \(\frac{d}{d x}\) [x log cos y]
⇒ y \(\frac{d}{d x}\) log cos x + log cos x \(\frac{d y}{d x}\) = x \(\frac{d}{d x}\) log cos y + log cos y . 1
⇒ \(\frac{y}{cos x}\) \(\frac{d}{d x}\) (cos x) + log cos x \(\frac{d y}{d x}\) = \(\frac{x}{cos y}\) \(\frac{d}{d x}\) (cos y) + log cos y
⇒ – y tan x + log cos x \(\frac{d y}{d x}\) = – x tan y \(\frac{d y}{d x}\) + log cos y
⇒ \(\frac{d y}{d x}\) (log cos x + x tan y) = log cos y + y tan x
∴ \(\frac{\log \cos y+y \tan x}{\log \cos x+x \tan y}\)

Question 11.
If y = x^{xx ………… ∞}, prove that \(\frac{d y}{d x}\) = \(\frac{y^2}{x(1-y \log x)}\).
Solution:
Given y = x^y …………(1)
Taking logarithm on both sides, we have ;
log y = y log x ;
\(\frac{1}{y} \frac{d y}{d x}=\frac{y}{x}+\log x \frac{d y}{d x}\)
⇒ \(\left(\frac{1}{y}-\log x\right) \frac{d y}{d x}=\frac{y}{x}\)
⇒ \(\left(\frac{1-y \log x}{y}\right) \frac{d y}{d x}=\frac{y}{x}\)
⇒ \(\frac{d y}{d x}=\frac{y^2}{x(1-y \log x)}\)

Exercise 5.9 (old)

Evaluate the following (1 to 5) limits (if they exist) :

Question 1.
(i) \(\ {Lt}_{x \rightarrow \infty} \frac{1}{(x-3)^2}\)
(ii) \(\ {Lt}_{x \rightarrow \infty} \frac{1}{(1-x)^2}\)
Solution:
(i) Let f(x) = \(\frac{1}{(x-3)^2}\) ;
D_f = R – {3}
as x Lakes positive values only and successive values of x go on increasing and becomes greater than any pre-assigned real number. however large it may be.
Then we say that x tends (goes) to infinity i.e. x → ∞
as x → ∞, x – 3 → ∞
(x – 3)² → ∞
i.e. \(\frac{1}{(x-3)^2}\) → 0
∴ \(\ {Lt}_{x \rightarrow 0} \frac{1}{(x-3)^2}\) = 0

(ii) Let f(x) = \(\frac{1}{(1-x)^2}\) ;
D_f = R – {1}
as x takes positive values only and successive values of x go on increasing and becomes greater than any pre assigned real number, however large it may be. Then we say that x → ∞
⇒ (1 – x)² → + ∞
⇒ \(\frac{1}{(1-x)^2}\) → 0
∴ \(\ {Lt}_{x \rightarrow \infty} \frac{1}{(1-x)^2}\) = 0.

Question 2.
(i) \(\underset{x \rightarrow \infty}{\ {Lt}}\) cos x
(ii) \(\underset{x \rightarrow 0}{\mathbf{L t}}\) cos \(\frac{1}{x}\)
Solution:
(i) Let cos x = f(x), D_f = R
When x = (2n + 1) \(\frac{\pi}{2}\) (n ∈ N)
Then cos x = 0, even for large values of n.
When x = 2nπ (n ∈ N)
Then cos x = 1, even for large values of n
When x = (2n+ 1) π (n ∈ N)
Then cos x = – 1, even for large values of n
Thus we observe that, cos x oscillates between – 1 and 1 as x → ∞
∴ \(\underset{x \rightarrow \infty}{\ {Lt}}\) cos x does not exists.

(ii) Let f(x) = cos \(\frac{1}{x}\) ; D_f = R – {0}
When x = \(\frac{1}{(2 n+1) \frac{\pi}{2}}\), n ∈ N and n is large
then x → 0 and
cos \(\frac{1}{x}\) = cos (2n + 1) \(\frac{\pi}{2}\) = 0
When x = \(\frac{1}{2 n \pi}\), n ∈ N and n is large then x → 0
∴ cos \(\frac{1}{x}\) = cos 2nπ = 1
When x = \(\frac{1}{2 n \pi+\pi}\), n ∈ N, n is large then x → 0
∴ cos \(\frac{1}{x}\) = cos (2n + 1) π = – 1
Thus, cos \(\frac{1}{x}\) oscillates between – 1 and 1,
as x → 0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) cos \(\frac{1}{x}\) does not exist.

Question 3.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^{-x}-1}{x}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}}{x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^{-x}-1}{x}\) = – \(\ {Lt}_{x \rightarrow 0} \frac{e^{-x}-1}{-x}\)
= – log e
= – 1
[∵ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{e^{-x}-1}{x}\) = log e = 1]

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}}{x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x-1-\left(e^{-x}-1\right)}{x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x-1}{x}-\ {Lt}_{x \rightarrow 0} \frac{e^{-x}-1}{x}\)
= log e + \(\ {Lt}_{x \rightarrow 0} \frac{e^{-x}-1}{-x}\)
= log e + log even= 1 + 1 = 2

Question 4.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{7^x-1}{\tan x}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{3^x-1}{\sqrt{2+x}-\sqrt{2}}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{7^x-1}{\tan x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{7^x-1}{x} \times \frac{x}{\tan x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{7^x-1}{x} \cdot \ {Lt}_{x \rightarrow 0} \frac{x}{\tan x}\)
= log 7 . 1 = log 7
[∵ \(\underset{\theta \rightarrow 0}{\ {Lt}} \frac{\theta}{\tan \theta}\) = 1]

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{3^x-1}{\sqrt{2+x}-\sqrt{2}}\)

Question 5.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-x-1}{x}\)
(ii) \(\ {Lt}_{x \rightarrow 1} \frac{x-1}{\log x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-x-1}{x}\)
= \(\ {Lt}_{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)-\ {Lt}_{x \rightarrow 0} 1\)
[∵ \(\ {Lt}_{x \rightarrow 0} \frac{e^x-1}{x}\) = log e = 1]
= log e – 1
= 1 – 1 = 0

(ii) put x = 1 + h
as x → 1 ⇒ h → 0
∴ \(\ {Lt}_{x \rightarrow 1} \frac{x-1}{\log x}=\ {Lt}_{h \rightarrow 0} \frac{1+h-1}{\log (1+h)}\)
= \(\ {Lt}_{h \rightarrow 0} \frac{h}{\log (1+h)}\)
= \(\frac{1}{\ {Lt}_{h \rightarrow 0} \frac{1}{h} \log (1+h)}\)
= \(\frac{1}{1}\) = 1
[∵ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{1}{x}\) log (1 + x) = 1]

Question 6.
Examine the function f(x) = \(\left\{\begin{array}{cc}
e^{1 / x} & , x \neq 0 \\
1 & , x=0
\end{array}\right.\) for continuity at x = 0.
Solution:
L.H.L. = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) e^1/x = 0
[as x → 0^–
⇒ x < 0
∴ \(\frac{1}{x}\) → – ∞
Thus e^1/x → e^{– ∞} = 0]
R.H.L. = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) e^1/x → + ∞
[as x → 0⁺ ⇒ x > 0
∴ \(\frac{1}{x}\) → + ∞
∴ e^1/x → ∞]
∴ L.H.L. ≠ R.H.L.
Thus f(x) is discontinuous at x = 0.