Differentiation Class 12 OP Malhotra Exe-8B ISC Maths Solutions Ch-8 Solutions. In this article you would learn about applying the chain rule and outer inner rule. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.
Differentiation Class 12 OP Malhotra Exe-8B ISC Maths Solutions Ch-8
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 12th |
| Chapter-8 | Differentiation |
| Writer | OP Malhotra |
| Exe-8(B) | applying the chain rule and outer inner rule |
Exercise- 8B
Differentiation Class 12 OP Malhotra Exe-8B Solution.
Que-1: (i) (5x + 7)10
(ii) 5/3x-1
(iii) √2-x^6
(iv) (2-x^5)^1/3
(v) (x²-5)^4
Sol: (i) Let y = (5x + 7)10
Diff both sides w.r.t x, we get
dy/dx = 10(5x + 7)10-1 dy/dx(5x + 7)
= 10(5x + 7)9 (5 x 1 + 0) = 50(5x + 7)9
(ii) 5/3x-1;
Diff both sides w.r.t x, we have
Que-2: (3x – x³ + 1)4
Sol: Let y = (3x – x³ + 1)4 ;
Diff both sides w.r.t x, we have
dy/dx = 4(3x – x³ + 1)³ dy/dx(3x – x³ + 1)
= 4(3x – x³ + 1)³ (3 – 3x²)
= 12(1 – x²) (3x – x³ + 1)³
Que-3: √x²+a²
Sol: Let y = (x²+a²)^1/2;
Diff both sides w.r.t x, we have
dy/dx=1/2(x²+a²)^(1/2-1) dy/dx(x²+a²)=1/2(x²+a²)^(-1/2) (2x+0)=x/√x²+a²
Que-4: 3/(a²-x²)²
Sol:
Que-5: √(ax²+bx+c)
Sol:
Que-6: (x² + 4)² (2x³ – 1)³
Sol: Let y = (x² + 4)² (2x³ – 1)³;
Diff both sides w.r.t x, we have
dy/dx=(x²+4)²d/dx(2x³-1)³+(2x³-1)d/dx(x²+4)²
=(x²+4)²3(2x³-1)²d/dx(2x³-1)+(2x³-1)2(x²+4)d/dx(x²+4)
=3(x²+4)²(2x³-1)²(6x²-0)+2(2x³-1)(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]
Que-7: x²/√4-x²
Sol:
Que-8: (x-1)√x²-2x+2
Sol:
Que-9: (x³-1/2x³+1)^4
Sol:
Que-10: (a+x/a-x)^(3/2)
Sol:
Que-11: a²+x²/√a²-x²
Sol:
Que-12: √1-x/2+x
Sol:
Que-13: √a+x – √a-x / √a+x + √a-x
Sol:
Que-14: √x²+1 + √x²-1 / √x²+1 – √x²-1
Sol:
Que-15: √1+√x
Sol:
Que-16: ((1+x²)^4)^1/3
Sol: Let y =[(1+x²)^4]^1/3= (1+x²)^4/3 ;
Diff both sides w.r.t x, we get
dy/dx=4/3(1+x²)^(4/3-1) d/dx(1+x²)=8x/3(1+x²)^(1/3)
Que-17: (i) sin x³
(ii) cos3x
(iii) tan √x
Sol: (i) Let y = sin x³ ; Diff both sides w.r.t x, we get
dy/dx=d/dx(sin x³)= cos x³ d/dx(x³)=3x² Cos x³
(if) Let y = cos³x = (cos x)³ ; Diff both sides w.r.t x, we have
dy/dx=3(cos x)^(3-1) d/dx(cos x)
=3cos²x (-sin x)
(iii) Let y = tan √x; Diff both sides w.r.t x, we have
Que-18: Sin 3x Cos 5x
Sol: Let y = sin 3x cos 5x
= 2 [2sin 3x cos 5x]
[∵ 2 sinA cos B = sin(A + B) + sin(A – B)]
⇒ y = 2 [sin 8x – sin 2x] ;
Diff both sides w.r.t x; we have
1/2 [8cos 8x – 2 cos 2x]
= 4 cos 8x – cos 2x
Que-19: Cos (Sin x²)
Sol: Let y = cos (sin x²) ;
Diff both sides w.r.t x, we get
∴ dy/dx =-sin(sin x²) d/dx sin x²
= – sin(sin x²) cos x² d/dx x²
= – 2x sin(sin x²) cos x²
Que-20: Cos² x³
Sol: Let y = cos²x³ = (cosx³)²
∴ dy/dx=d/dx(cosx³)²=2(cosx³)^(2-1) d/dx(cos x³)
= 2cosx³ (- sin x³). d/dx x³
= – 3x² sin(2x³)
[∵ 2sin θ cos θ sin2θ]
Que-21: √a+√a+x
Sol:
Que-22: |x²-7|
Sol:
Que-23: y=u-1/u+1 , u=√x
Sol:
Que-24: y=t³+4 , t=x²+2x
Sol: Given y = t³ + 4 … (1)
& t = x² + 2x … (2)
Diff (1) w.r.t t, we have
dy/dx=3t² ;
DifF (2) w.r.t x, we have dt
dy/dx = 2x + 2
∴ dy/dx=dy/dt×dt/dx=3t²(2x +2)
= 3(x² + 2x)²(2x + 2) [using eqn. (1)]
⇒ dy/dx
= 3x² (x + 2)² x 2(x + 1)
= 6x²(x + 1)(x + 2)²
Que-25: y=u²-1/u²+1 and u =(x²+2)^(1/3)
Sol:
Que-26: Find dy/dx if y = u4 , u =1/√v and v = 5x² + 2x + 6.
Sol: Given y = u4 … (1)
u = 1/√v (2)
& v = 5x² + 2x + 6 … (3)
DifF. (1) w.r.t w; we have
dy/du = 4u³ … (4)
DifF. (2) w.r.t w; we have
dy/dv=-1/2v^(3/2) … (5)
DifF. (3) w.r.t w; we have
dy/dx = 10x + 2 … (6)
Thus, 1/2
= 4u³(-1/2v^(3/2))(10x+2) [using eqn (4), (5) & (6)]
= -4u³/v^(3/2) (5x+1)
Que-27: Given that y=5x/(1-x)^(2/3) +cos²(2x+1) , show that dy/dx=5/3(1-x)^(-5/3) (3-x) -2 sin(4x+2)
Sol:
Que-28: If y=1/2 log(1-cos 2x /1+cos 2x) , prove that dy/dx=2 cosec x.
Sol:
Que-29: Given y=(√x+1)(x²-√x)/x√x+x+√x +1/15(3 cos²x-5)cos³x , show that dy/dx=1+ sin³x cos²x.
Sol:
Que-30: Given y=(3x-1)²+(2x-1)³,find that dy/dx and the points on the curve for which dy/dx=0.
Sol: Given y = (3x – 1)² + (2x – 1)³ …(1)
Diff both sides w.r.t x, we have
dy/dx = 2(3x – 1) d/dx (3x – 1) + 3(2x – 1)²d/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. dy/dx = 0
⇒ 6(4x – 1)x = 0
⇒ x =0,1/4
when x = 0 ∴ from (1) ; y = (0 – 1)² + (0 – 1)³ = 1 – 1 = 0
when x = 4 ∴ from (1) ; y =(3/4-1)²+(1/2-1)³=1/16+1/8=3/16
Thus, required points on given curve are (0, 0) & (1/4,3/16).
–: End Differentiation Class 12 OP Malhotra Exe-8B ISC Math Ch-8 Solution :–
Return to :- OP Malhotra ISC Class-12 S Chand Publication Maths Solutions
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