Samacheer Kalvi 11th Maths Guide Chapter 10 கணங்கள், தொடர்புகள் மற்றும் சார்புகள் Ex 10.3

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 10 கணங்கள், தொடர்புகள் மற்றும் naசார்புகள் Ex 10.3 Textbook Questions and Answers, Notes.

TN Board 11th Maths Solutions Chapter 10 கணங்கள், தொடர்புகள் மற்றும் சார்புகள் Ex 10.3

கீழ்க்காணும் சார்புகளுக்கு வகைக்கெழுக்களைக் காண்க.

Question 1.
y = (x² + 4x + 6)⁵
தீர்வு :
y = (x² + 4x + 6)⁵
u = x² + 4x + 6
⇒ y = u⁵ என்க..
∴ \(\frac{d u}{d x}\) = 2x + 4 மற்றும் \(\frac{d y}{d u}\) = 5u⁴
∴ \(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\) = 5u⁴ (2x + 4)
= 5 (2x + 4) (x² + 4x + 6)⁴
[∵ u = x² + 4x + 6]

Question 2.
y = tan 3x
தீர்வு :
y = tan 3x
u = 3x
⇒ y = tan u என்க
\(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= sec² x × 3 = sec² u = 3sec²(3x) [∵ u = 3x]

Question 3.
y = cos(tan x)
தீர்வு :
y = cos(tan x)
u = tan x ⇒ y = cos u என்க
\(\frac{d u}{d x}\) = sec² x மற்றும் \(\frac{d y}{d u}\) = -sin u
∴ \(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= -sin u × sec² x
= -sin(tan x) . sec² x [∵ u = tan x]

Question 4.
y = \(\sqrt[3]{1+x^{3}}\)
தீர்வு :
y = \(\sqrt[3]{1+x^{3}}\) = \(\left(1+x^{3}\right)^{\frac{1}{3}}\)
u = 1 + x³ ⇒ y = u^{\(\frac{1}{3}\)} என்க
∴ \(\frac{d u}{d x}\) = 3x² மற்றும் \(\frac{d y}{d u}\) = \(\frac{1}{3}\) \(u^{\frac{1}{3}-1}\)
= \(\frac{1}{3}\) \(u^{\frac{-2}{3}}\)

∴ \(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= \(\frac{1}{3} u^{\frac{-2}{3}} \cdot\left(3 x^{2}\right)\)
= \(\frac{1}{3}\) \(\left(1+x^{3}\right)^{\frac{-2}{3}}\) × 3x²
= x²(1 + x³)^{\(\frac{-2}{3}\)} 

Question 5.
y = \(e^{\sqrt{x}}\)
தீர்வு :
y = \(e^{\sqrt{x}}\)
u = √x ⇒ y = e^u என்க
∴ \(\frac{d u}{d x}\) = \(\frac{1}{2}\) x^{–\(\frac{1}{2}\) – 1};

\(\frac{d y}{d u}\) = e^u

⇒ \(\frac{d y}{d u}\) = \(\frac{1}{2}\) x^{\(-\frac{1}{2}\)} = \(\frac{1}{2 \sqrt{x}}\)

∴ \(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= e^u × \(\frac{1}{2 \sqrt{x}}\)
= \(\frac{e^{\sqrt{x}}}{2 \sqrt{x}}\) [∵ u = √x]

Question 6.
y = sin(e^x)
தீர்வு :
y = sin(e^x)
u = e^x
⇒ y = sin u என்க
∴ \(\frac{d u}{d x}\) = e^x மற்றும் \(\frac{d y}{d u}\) = cos u
\(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= cos u × e^x = e^x cos(e^x) [∵ u = e^x]

Question 7.
F(x) = (x³ + 4x)⁷
தீர்வு :
F(x) = (x³ + 4x)⁷
u = x³ + 4x
⇒ F(x) = u⁷ என்க
∴ \(\frac{d u}{d x}\) = 3x² + 4 மற்றும் F'(u) = 7u⁶
∴ F'(x) = F'(u) . \(\frac{d u}{d x}\) = 7 . u⁶(3x² + 4)
= 7 . (x³ + 4x)⁶ (3x² +4)
F'(x) = 7(3x² + 4) (x³ + 4x)⁶

Question 8.
h(t) = (t – \(\frac{1}{t}\))^{\(\frac{3}{2}\)}
தீர்வு :
h(t) = (t – \(\frac{1}{t}\))^{\(\frac{3}{2}\)}
u = t – \(\frac{1}{t}\)
⇒ h(u) = u^{\(\frac{3}{2}\)} என்க
∴ \(\frac{d u}{d t}\) = 1 + \(\frac{1}{t^{2}}\) மற்றும்
h'(u) = \(\frac{3}{2} u^{\frac{3}{2}-1}=\frac{3}{2} u^{\frac{1}{2}}\)
h'(t) = h'(u) . \(\frac{d u}{d t}\)
= \(\frac{3}{2} u^{\frac{1}{2}}\left(1+\frac{1}{t^{2}}\right)\) [d(1 + \(\frac{1}{t}\)) = –\(\frac{1}{t^{2}}\)]

h'(t) = \(\frac{3}{2}\left(t-\frac{1}{t}\right)^{\frac{1}{2}}\left(1+\frac{1}{t^{2}}\right)\) [∵ u = t – \(\frac{1}{t}\)]

Question 9.
f(t) = \(\sqrt[3]{1+\tan t}\)
தீர்வு :
f(t) = \(\sqrt[3]{1+\tan t}\) = (1 + tan t)^{\(\frac{1}{3}\)}
u = 1 + tan t
⇒ f(u) = u^{\(\frac{1}{3}\)} என்க
∴ \(\frac{d u}{d t}\) = sec² t மற்றும்
f'(u) = \(\frac{1}{3} u^{\frac{1}{3}-1}=\frac{1}{3} u^{-\frac{2}{3}}\)
∴ f'(t) = f'(u) . \(\frac{d u}{d t}\)

= \(\frac{1}{3} u^{-\frac{2}{3}} \sec ^{2} t\)

= \(\frac{1}{3} \sec ^{2} t(1+\tan t)^{-\frac{2}{3}}\) [∵ u = 1 + tan t]

Question 10.
y = cos(a³ + x³)
தீர்வு :
y = cos(a³ + x³)
u = a³ + x³ ⇒ y = cos u என்க

∴ \(\frac{d u}{d x}\) = 0 + 3x² = 3x² மற்றும் \(\frac{d y}{d u}\) = -sin u

∴ \(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= – sin u × 3x²
= -3x² sin u

∴ \(\frac{d y}{d u}\) = – 3x² . sin((a³ + x³)

Question 11.
y = e^-mx
தீர்வு :
y = e^-mx
u = -mx
⇒ y = e^u என்க
∴ \(\frac{d u}{d x}\) = -m மற்றும் \(\frac{d y}{d u}\) = e^u

∴ \(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d y}{d u}\)
= e^u × (-m) = -me^u
= -m e^-mx [∵ u = -mx]
= -my [∵ y = e^-mx]

Question 12.
y = 4 sec 5x
தீர்வு :
y = 4 sec 5x
u = 5x ⇒ y = 4 sec u என்க
∴ \(\frac{d u}{d x}\) = 5 மற்றும் \(\frac{d y}{d u}\) = 4sec u tan u
∴ \(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= 4 sec u tan u × 5
= 20 sec 5x tan 5x [∵ u = 5x]

Question 13.
y = (2x – 5)⁴ (8x² – 5)^-3
தீர்வு :
y = (2x – 5)⁴ (8x² – 5)^-3
u = 2x – 5, v = 8x² – 5 என்க.
⇒ \(\frac{d u}{d x}\) = 2 மற்றும் \(\frac{d v}{d x}\) = 16 x
⇒ y = u⁴ v^-3
∴ \(\frac{d y}{d u}\) = u⁴ . \(\frac{d}{d x}\)(v^-3) + (v^-3). \(\frac{d}{d x}\)(u⁴) (வகுத்தல் விதி)

= u⁴ . (-3) v^-3-1 \(\frac{d v}{d x}\) + v^-3 . 4u³ . \(\frac{d u}{d x}\)
= -3u⁴v^-4 . \(\frac{d v}{d x}\) + 4u³ . v^-3 \(\frac{d u}{d x}\)

= 3u⁴ v^-4 . \(\frac{d v}{d x}\) + 4u³v^-3 . \(\frac{d u}{d x}\)

= -3u⁴v^-4(16x) + 4u³v^-3(2)
= -48 x u⁴v^-4 + 8 u³v^-3
=

Question 14.
y = (x² + 1)\(\sqrt[3]{x^{2}+2}\)
தீர்வு :
y = (x² + 1)\(\sqrt[3]{x^{2}+2}\)
u = x² + 1 மற்றும் v = x² + 2 என்க.
⇒ \(\frac{d u}{d x}\) = 2x மற்றும் \(\frac{d v}{d x}\) = 2x
∴ y = u . v^{\(\frac{1}{3}\)}
∴ \(\frac{d y}{d u}\) = u . \(\frac{d}{d x}\)(v^{\(\frac{1}{3}\)}) + v^{\(\frac{1}{3}\)} . \(\frac{d}{d x}\)(u)

= u . \(\frac{1}{3}\) (v^{\(\frac{1}{3}\) – 1}) . \(\frac{d v}{d x}\) + v^{\(\frac{1}{3}\)} . \(\frac{d u}{d x}\)

= \(\frac{\left(x^{2}+1\right)}{3}\) . (x² + 2)^{\(-\frac{2}{3}\)}(2x) + (x² + 2)^{\(\frac{1}{3}\)} . 2x

= \(\frac{2 x\left(x^{2}+1\right)}{3\left(x^{2}+2\right)^{\frac{2}{3}}}\) + (2x) (x² + 2)^{\(\frac{1}{3}\)}

= \(\frac{2 x^{3}+2 x+2 x\left(x^{2}+2\right)^{\frac{1}{3}} \cdot 3\left(x^{2}+2\right)^{\frac{2}{3}}}{3\left(x^{2}+2\right)^{\frac{2}{3}}}\)

= \(\frac{2 x^{3}+2 x+6 x\left(x^{2}+2\right)^{1}}{3\left(x^{2}+2\right)^{\frac{2}{3}}}\)

= \(\frac{2 x^{3}+2 x+6 x^{3}+12 x}{3\left(x^{2}+2\right)^{\frac{2}{3}}}\)

= \(\frac{8 x^{3}+1}{3\left(x^{2}+2\right)^{\frac{2}{3}}}\)

Question 15.
y = xe^-x2
தீர்வு :
y = xe^-x2
u = x மற்றும் v = -x² என்க
⇒ \(\frac{d u}{d x}\) = 1 மற்றும் \(\frac{d v}{d x}\) = -2x
∴ y = u . e^y
∴ \(\frac{d y}{d u}\) = u . e^v \(\frac{d v}{d x}\) + e^v \(\frac{d u}{d x}\)
= x . e^-x2(-2x) + e^-x2(1)
= -2x²e^-x2 + e^-x2
= e^-x2(1 – 2x²)

Question 16.
s(t) = \(\sqrt[4]{\frac{t^{3}+1}{t^{3}-1}}\)
தீர்வு :
s(t) = \(\sqrt[4]{\frac{t^{3}+1}{t^{3}-1}}\)
= (t³ + 1)^{\(\frac{1}{4}\)} (t³ – 1)^{\(-\frac{1}{4}\)}
u = t³ + 1 மற்றும் v = t³ – 1 என்க
⇒ \(\frac{d u}{d t}\) = 3t² மற்றும்
∴ s(t) = u^{\(\frac{1}{4}\)} . v^{\(-\frac{1}{4}\)}
s'(t) = u^{\(\frac{1}{4}\)}(-\(\frac{1}{4}\))v^{\(-\frac{1}{4}\) – 1} . \(\frac{d v}{d t}\) + v^{\(-\frac{1}{4}\)} . \(\frac{1}{4}\) u^{\(\frac{1}{4}\) – 1} . \(\frac{d u}{d t}\)

=

Question 17.
f(x) = \(\frac{x}{\sqrt{7-3 x}}\)
தீர்வு :
y = \(\frac{x}{\sqrt{7-3 x}}\)

Question 18.
y = tan(cos x)
தீர்வு :
y = tan(cos x)
u = cos x ⇒ y = tan u என்க
∴ \(\frac{d u}{d x}\) = – sin x மற்றும் \(\frac{d u}{d x}\) = sec² u

∴ \(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= sec² u(- sin x)
= -sin xsec²(cos x) [∵ u = cos x]

Question 19.
y = \(\frac{\sin ^{2} x}{\cos x}\)
தீர்வு :

= sin x(\(\frac{1}{\cos ^{2} x}+\frac{\cos ^{2} x}{\cos ^{2} x}\))
= sin x(sec² x + 1)

Question 20.
y = \(5^{\frac{-1}{x}}\)
தீர்வு :
y = \(5^{\frac{-1}{x}}\)
u = –\(\frac{1}{x}\), y = 5^u என்க
∴ \(\frac{d u}{d x}\) = \(\frac{1}{x^{2}}\), \(\frac{d y}{d u}\) = 5^u log 5
∴ \(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= 5^u (log 5) \(\frac{1}{x^{2}}\)
= \(\frac{5^{-\frac{1}{x}}}{x^{2}}\) log 5

Question 21.
y = \(\sqrt{1+2 \tan x}\)
தீர்வு :
y = (1 + 2tan x)^{\(\frac{1}{2}\)}
u = 1 + 2tan x என்க y = u^{\(\frac{1}{2}\)}
\(\frac{d u}{d x}\) = 2 sec² x மற்றும் \(\frac{d y}{d u}\) = \(\frac{1}{2}\) u^{–\(\frac{1}{2}\)} = \(\frac{1}{2 \sqrt{u}}\)

\(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= \(\frac{1}{2 \sqrt{u}}\) × 2 sec² x = \(\frac{\sec ^{2} x}{\sqrt{1+2 tan u}}\)

Question 22.
y = sin³ x + cos³ x
தீர்வு :
\(\frac{d y}{d u}\) = \(\frac{d}{d x}\)(sin³ x) . \(\frac{d}{d x}\) (cos³ x)
= 3 sin² x . \(\frac{d}{d x}\)(sin x) + 3 cos² x . \(\frac{d}{d x}\)(cos x)
= 3 sin² x cos x + 3 cos² x(-sin x)
= 3 sin² x cos x + 3 sin x cos² x
= 3 sin x cos x(sin x – cos x)

Question 23.
y = sin²(cos kx)
தீர்வு :
y = sin²(cos kx)
u = cos kx ⇒ y = sin² u
∴ \(\frac{d u}{d x}\) = -k sin kx மற்றும் \(\frac{d y}{d u}\)
= 2 sin u . \(\frac{d}{d u}\) (sin u) = 2 sin u cos u
\(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= 2 sin u cos u × (-k sin kx)
= – k sin 2u sin kx [∵ u = cos kx]
= – k sin 2(cos kx) . sin kx
= -k sin kx . sin 2(cos kx)

Question 24.
y = (1 + cos2 x)⁶
தீர்வு :
y = (1 + cos2 x)⁶
u = 1 + cos² x ⇒ y = u⁶ என்க
∴ \(\frac{d u}{d x}\) = 2 cos x(-sin x)
\(\frac{d y}{d u}\) = 2 cos x(-sin x)
\(\frac{d y}{d u}\) = 6u⁵ = -2 sin x cos x = – sin 2x
\(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= 6u⁵ (- sin 2x)
= -6(1 + cos² x)⁵ sin 2x
= -6 sin 2x(1 + cos² x)⁵

Question 25.
y = \(\frac{e^{3 x}}{1+e^{x}}\)
தீர்வு :
y = \(\frac{e^{3 x}}{1+e^{x}}\)
வகுத்தல் விதியைப் பயன்படுத்தி

Question 26.
y = \(\sqrt{x+\sqrt{x}}\)
தீர்வு :
y = \(\sqrt{x+\sqrt{x}}\)
y² = x + √x
∴ 2y \(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x) + \(\frac{d}{d x}\)(√x)

⇒ 2y \(\frac{d y}{d u}\) = 1 + \(\frac{1}{2 \sqrt{x}}\)
= \(\frac{2 \sqrt{x}+1}{2 \sqrt{x}}\)

⇒ \(\frac{d y}{d u}\) = \(\frac{1}{2 y}\left[\frac{2 \sqrt{x}+1}{2 \sqrt{x}}\right]\)
= \(\frac{1}{2 \sqrt{x+\sqrt{x}}}\left[\frac{2 \sqrt{x}+1}{2 \sqrt{x}}\right]\)

⇒ \(\frac{d y}{d u}\) = \(\frac{2 \sqrt{x}+1}{4 \sqrt{x} \cdot \sqrt{x+\sqrt{x}}}\)

Question 27.
y = e^{x cos x}
தீர்வு :
y = e^{x cos x}
u = x cos x ⇒ y = e^u
∴ \(\frac{d y}{d u}\) = x(-sin x) + cos x
= -x sin x + cos x மற்றும் \(\frac{d y}{d u}\) = e^u

∴ \(\frac{d y}{d u}\) = \(\frac{d y}{d u}\) × \(\frac{d u}{d x}\)
= e^u(-x sin x + cos x)
= e^{x cos x}(cos x – x sin x)

Question 28.
y = \(\sqrt{x+\sqrt{x+\sqrt{x}}}\)
தீர்வு :
y = \(\sqrt{x+\sqrt{x+\sqrt{x}}}\)
இருபுறமும் வர்க்கப்படுத்த,
y² = x + \(\sqrt{x+\sqrt{x}}\)
∴ 2y \(\frac{d y}{d u}\) = \(\frac{d}{d x}\) (x) + \(\frac{d}{d x}\) (\(\sqrt{x+\sqrt{x}}\))

Question 29.
y = sin(tan(\(\sqrt{\sin x}\)))
தீர்வு :
y = sin(tan(\(\sqrt{\sin x}\)))
u = \(\sqrt{\sin x}\)) மற்றும்
v = tan u ⇒ y = sin v என்க

⇒ \(\frac{d u}{d x}\) = \(\frac{1}{2}(\sin x)^{-1 / 2}\) . cos x;

\(\frac{d v}{d u}\) = sec² u; \(\frac{d y}{d v}\) = cos v = \(\frac{\cos x}{2 \sqrt{\sin x}}\)

∴ \(\frac{d y}{d u}\) = \(\frac{d y}{d v}\) × \(\frac{d v}{d u}\) × \(\frac{d u}{d x}\)
= (cos v) (sec² u) \(\frac{\cos x}{2 \sqrt{\sin x}}\)
= cos(tan u) . sec² (\(\sqrt{\sin x}\)) . \(\frac{\cos x}{2 \sqrt{\sin x}}\)
= cos(tan(\(\sqrt{\sin x}\))) . sec² (\(\sqrt{\sin x}\)) . \(\frac{\cos x}{2 \sqrt{\sin x}}\)

Question 30.
y = \(\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
தீர்வு :
y = \(\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
t = \(\frac{1-x^{2}}{1+x^{2}}\)

⇒ \(\frac{d t}{d x}\) = \(\frac{\left(1+x^{2}\right)(-2 x)-\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}(2 x)\)

= \(\frac{-2 x-2 x^{3}-2 x+2 x^{5}}{\left(1+x^{2}\right)^{2}}=\frac{4 x}{\left(1+x^{2}\right)^{2}}\) ………… “(1)

y = sin^-1 t
∴ \(\frac{d y}{d x}\) = \(\frac{d}{d t}\) (sin^-1 t) . \(\frac{d t}{d x}\)

= \(\frac{1}{\sqrt{1-t^{2}}}\left(\frac{4 x}{\left(1+x^{2}\right)^{2}}\right.\)

\(\sqrt{1-t^{2}}\) = \(\sqrt{1-\frac{\left(1-x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}}}\)

= \(\sqrt{\frac{\left(1+x^{2}\right)^{2}-\left(1-x^{2}\right)^{2}}{\left(1+x^{ 2}\right)^{2}}}\)

= \(\sqrt{\frac{1+x^{4}+2 x^{2}-\left(1+x^{4}-2 x^{2}\right)}{\left(1+x^{2}\right)^{2}}}\)

= \(\sqrt{\frac{4 x^{2}}{\left(1+x^{2}\right)^{2}}}=\frac{2 x}{1+x^{2}}\) ………..(2)
(2)ஐ (1)-ல் பிரதியிட,
\(\frac{d y}{d u}\) = \(\frac{1+x^{2}}{2 x}\left(\frac{-4 x}{\left(1+x^{2}\right)^{2}}\right)\) = \(\frac{-2}{1+x^{2}}\)