OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(b)

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S Chand Class 12 ICSE Maths Solutions Chapter 8 Differentiation Ex 8(b)

Question 1.
(i) (5x + 7)¹⁰
(ii) \(\frac{5}{3 x-1}\)
(iii) \(\sqrt{2-x^6}\)
(iv) \(\sqrt[3]{2-x^5}\)
(v) \(\left(x^2-5\right)^4\)
Solution:
(i) Let y = (5x + 7)¹⁰
Diff both sides w.r.t x, we get
\(\frac { dy }{ dx }\) = 10(5x + 7)^10-1 \(\frac { dy }{ dx }\)(5x + 7)
= 10(5x + 7)⁹ (5 x 1 + 0) = 50(5x + 7)⁹

(ii) \(\frac{5}{3 x-1}\);
Diff both sides w.r.t x, we have

Question 2.
(3x – x³ + 1)⁴
Solution:
Let y = (3x – x³ + 1)⁴ ;
Diff both sides w.r.t x, we have
\(\frac { dy }{ dx }\) = 4(3x – x³ + 1)³ \(\frac { dy }{ dx }\)(3x – x³ + 1)
= 4(3x – x³ + 1)³ (3 – 3x²)
= 12(1 – x²) (3x – x³ + 1)³

Question 3.
\(\sqrt{x^2+a^2}\)
Solution:
Let y = \(\left(x^2+a^2\right)^{\frac{1}{2}}\);
Diff both sides w.r.t x, we have
∴ \(\frac{d y}{d x}=\frac{1}{2}\left(x^2+a^2\right)^{\frac{1}{2}-1} \frac{d}{d x}\left(x^2+a^2\right) \Rightarrow \frac{d y}{d x}=\frac{1}{2}\left(x^2+a^2\right)^{\frac{-1}{2}}(2 x+0)=\frac{x}{\sqrt{x^2+a^2}}\)

Question 4.
\(\frac{3}{\left(a^2-x^2\right)^2}\)
Solution:

Question 5.
\(\sqrt{\left(a x^2+b x+c\right)}\)
Solution:

Question 6.
(x² + 4)² (2x³ – 1)³
Solution:
Let y = (x² + 4)² (2x³ – 1)³;
Diff both sides w.r.t x, we have
\(\frac{d y}{d x}=\left(x^2+4\right)^2 \frac{d}{d x}\left(2 x^3-1\right)^3+\left(2 x^3-1\right)^3 \frac{d}{d x}\left(x^2+4\right)^2\)
= \(\left(x^2+4\right)^2 3\left(2 x^3-1\right)^2 \frac{d}{d x}\left(2 x^3-1\right)+\left(2 x^3-1\right) 2\left(x^2+4\right) \frac{d}{d x}\left(x^2+4\right)\)
= 3(x² + 4)² (2x³ – 1)² (6x² – 0) + (2x³ – 1)³ 2(x² + 4)2x
= (x² + 4)² (2x² – 1)² 2x[9x(x² + 4) + 2(2x³ – 1)]
= (x² + 4) (2x³ – 1)² 2x[ 13x³ + 36x – 2]

Question 7.
\(\frac{x^2}{\sqrt{4-x^2}}\)
Solution:

Question 8.
\((x-1) \sqrt{x^2-2 x+2}\)
Solution:

Question 9.
\(\left(\frac{x^3-1}{2 x^3+1}\right)^4\)
Solution:

Question 10.
\(\left(\frac{a+x}{a-x}\right)^{3 / 2}\)
Solution:

Question 11.
\(\frac{a^2+x^2}{\sqrt{a^2-x^2}}\)
Solution:

Question 12.
\(\sqrt{\frac{1-x}{2+x}}\)
Solution:

Question 13.
\(\frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}\)
Solution:

Question 14.
\(\frac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}}\)
Solution:

Question 15.
\(\sqrt{1+\sqrt{x}}\)
Solution:

Question 16.
\(\sqrt[3]{\left(1+x^2\right)^4}\)
Solution:
Let y = \(\sqrt[3]{\left(1+x^2\right)^4}\) = \(\left[\left(1+x^2\right)^4\right]^{1 / 3}=\left(1+x^2\right)^{4 / 3}\) ; Diff both sides w.r.t x, we get
\(\frac{d y}{d x}=\frac{4}{3}\left(1+x^2\right)^{\frac{4}{3}-1} \frac{d}{d x}\left(1+x^2\right)=\frac{8 x}{3}\left(1+x^2\right)^{\frac{1}{3}}\)

Question 17.
(i) sin x³
(ii) cos3x
(iii) tan \(\sqrt{x}\)
Solution:
(i) Let y = sin x³ ; Diff both sides w.r.t x, we get
\(\frac{d y}{d x}=\frac{d}{d x} \sin x^3=\cos x^3 \frac{d}{d x} x^3\) = 3x² cos x³

(if) Let y = cos³x = (cos x)³ ; Diff both sides w.r.t x, we have
\(\frac { dy }{ dx }\) = 3(cos x)^3-1 \(\frac { d }{ dx }\) cos x
= 3 cos²x (- sin x)

(iii) Let y = tan \(\sqrt{x}\) ; Diff both sides w.r.t x, we have

Question 18.
sin 3x cos 5x
Solution:
Let y = sin 3x cos 5x
= \(\frac { 1 }{ 2 }\) [2sin 3x cos 5x]
[∵ 2 sinA cos B = sin(A + B) + sin(A – B)]
⇒ y = \(\frac { 1 }{ 2 }\) [sin 8x – sin 2x] ;
Diff both sides w.r.t x; we have
\(\frac { 1 }{ 2 }\) [8cos 8x – 2 cos 2x]
= 4 cos 8x – cos 2x

Question 19.
cos (sin x²)
Solution:
Let y = cos (sin x²) ;
Diff both sides w.r.t x, we get
∴ \(\frac{d y}{d x}=-\sin \left(\sin x^2\right) \frac{d}{d x} \sin x^2\)
= – sin(sin x²) cos x²\(\frac { d }{ dx }\)x²
= – 2x sin(sin x²) cos x²

Question 20.
cos2x³
Solution:
Let y = cos²x³ = (cosx³)²
∴ \(\frac{d y}{d x}=\frac{d}{d x}\left(\cos x^3\right)^2=2 \cos x^3 \frac{d}{d x} \cos x^3\)
= 2cosx³ (- sin x³). \(\frac { d }{ dx }\) x³
= – 3x² sin(2x³)
[∵ 2sin θ cos θ sin2θ]

Question 21.
\(\sqrt{a+\sqrt{a+x}}\)
Solution:

Question 22.
\(\left|x^2-7\right|\)
Solution:

Question 23.
y = \(\frac{u-1}{u+1}, u=\sqrt{x}\)
Solution:

Question 24.
y = t³ + 4, t = x² + 2x
Solution:
Given y = t³ + 4 … (1)
& t = x² + 2x … (2)
Diff (1) w.r.t t, we have
\(\frac { dy }{ dx }\) = 3t² ;
DifF (2) w.r.t x, we have dt
\(\frac { dy }{ dx }\) = 2x + 2
∴ \(\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=3 t^2(2 x+2)\)
= 3(x² + 2x)²(2x + 2) [using eqn. (1)]
⇒ \(\frac { dy }{ dx }\)
= 3x² (x + 2)² x 2(x + 1)
= 6x²(x + 1)(x + 2)²

Question 25.
y = \(\frac{u^2-1}{u^2+1}\) and u = \(\sqrt[3]{x^2+2}\)
Solution:

Question 26.
Find \(\frac { dy }{ dx }\) if y = u⁴ , u = \(\frac{1}{\sqrt{v}}\) and v = 5x² + 2x + 6.
Solution:
Given y = u⁴ … (1)
u = \(\frac{1}{\sqrt{v}}\) … (2)
& v = 5x² + 2x + 6 … (3)
DifF. (1) w.r.t w; we have
\(\frac { dy }{ du }\) = 4u³ … (4)
DifF. (2) w.r.t w; we have
\(\frac{d u}{d v}=\frac{-1}{2 v^{3 / 2}}\) … (5)
DifF. (3) w.r.t w; we have
\(\frac { dy }{ dx }\) = 10x + 2 … (6)
Thus, \(\frac { 1 }{ 2 }\)
= \(4 u^3\left(\frac{-1}{2 v^{3 / 2}}\right)(10 x+2)\) [using eqn (4), (5) & (6)]
= \(\frac{-4 u^3}{v^{3 / 2}}(5 x+1)\)

Question 27.
Given that y = \(\frac{5 x}{(1-x)^{2 / 3}}+\cos ^2(2 x+1)\) + cos² (2x + 1), show that \(\frac{d y}{d x}=\frac{5}{3}(1-x)^{-5 / 3}(3-x)-2 \sin (4 x+2)\)
Solution:

Question 28.
If y = \(\frac{1}{2} \log \left(\frac{1-\cos 2 x}{1+\cos 2 x}\right)\) prove that \(\frac { dy }{ dx }\) = 2 cosec x.
Solution:

Question 29.
Given y = \(\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\) (3 cos² x – 5)cos³ x, show that \(\frac { dy }{ dx }\) = 1 + sin³ x cos² x
Solution:

Question 30.
Given y = (3x – 1)² + (2x – 1)³, find \(\frac { dy }{ dx }\) and the points on the curve for which \(\frac { dy }{ dx }\) = 0
Solution:
Given y = (3x – 1)² + (2x – 1)³ …(1)
Diff both sides w.r.t x, we have
\(\frac { dy }{ dx }\) = 2(3x – 1) \(\frac { d }{ dx }\) (3x – 1) + 3(2x – 1)² \(\frac { d }{ dx }\)(2x – 1)
= 2(3x – 1).3 + 3(2x – l)² x 2 = 6 [3x – 1 + (2x – 1)²]
= 6(4x² – x) – 6x(4x – 1)
Let (x, y) be any point on given curve s.t. \(\frac { dy }{ dx }\) = 0
⇒ 6(4x – 1)x = 0
⇒ x = 0, \(\frac { 1 }{ 4 }\)
when x = 0 ∴ from (1) ; y = (0 – 1)² + (0 – 1)³ = 1 – 1 = 0
when x = \(\frac { 1 }{ 4 }\) ∴ from (1) ; y = \(\left(\frac{3}{4}-1\right)^2+\left(\frac{1}{2}-1\right)^3=\frac{1}{16}+\frac{1}{8}=\frac{3}{16}\)
Thus, required points on given curve are (0, 0) & \(\left(\frac{1}{4}, \frac{3}{16}\right)\).