Differentiation Class 12 OP Malhotra Exe-8A ISC Maths Solutions Ch-8 Solutions. In this article you would learn about the basics of differentiation. 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-8A ISC Maths Solutions Ch-8
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 12th |
| Chapter-8 | Differentiation |
| Writer | OP Malhotra |
| Exe-8(A) | the basics of differentiation |
Exercise- 8A
Differentiation Class 12 OP Malhotra Exe-8A Solution.
Que-1:
(i) x5
(ii) 6x8
(iii) x3/4
(iv) 4√x
(v) 8x-3/4
(vi) √x³
(vii) 9/x
(viii) 7/x²
Sol: (i) Let y = x5,
Diff both sides w.r.t x, we have
dy/dx=d/dx(x^5)=5x^4
(ii) Let y = 6x8,
Diff both sides w.r.t x, we have
dy/dx=d/dx(6x8)=48x^7
(iii) Let y = x3/4
Diff both sides w.r.t x, we have
dy/dx=d/dx(x3/4)=3/4=3/4x^(-1/4)
(iv) Let y = 4√x
Diff both sides w.r.t x, we have
dy/dx=d/dx(4√x)=4d/dx(x^1/2)
=4×1/2x^(1/2-1)=2/√x
(v) Let y =8x^(-3/4);
Diff both sides w.r.t x, we have
dy/dx=8(-3/4)d/dx(x^(3/4-1)=-6x^(-7/4)
(vi) Let y = √x³
Diff both sides w.r.t x, we have
dy/dx=d/dx(x^(3/2))=3/2x^(3/2-1)=3/2√x
(vii) Let y = 9/x
Diff both sides w.r.t x, we have
dy/dx=9d/dx(1/x)=9d/dx(x^(-1))=9(-1)x^(-1-1)
(viii) Let y = 7/x² ;
Diff both sides w.r.t x, we have
dy/dx=7d/dx(x^(-2))=7(-2)x^(-2-1)=-14/x³
Que-2: (i) (2x + 3)5
(ii) (1 – x)4
(iii)√8-7x
(iv) (3x² + 5)9
Sol:
Que-3: (i) 2/x+1/√x
(ii) (2x – 1) (3x + 2)
(iii) x^4-2x+1/x²
(iv) 2x² (x + 1) + 2
(v) (2x+3)^5
Sol:
Que-4: (i) cos 7x
(ii) tan ax
(iii) sec 9x
(iv) sin x²
(v) cos √x
(vi) 2 cosec bx³
Sol: (i) Let y = cos 7x ;
Diff both sides w.r.t x, we have
dy/dx=d/dx(cos 7x)
= sin 7x d/dx(7x) = – 7sin 7x
(ii) Let y = tan ax ;
Diff both sides w.r.t x, we have
dy/dx=(sec²ax)d/dx(ax)=asec²ax
(iii) Let y = sec 9x ;
Diff both sides w.r.t x, we have
dy/dx=d/dx(sec9x)=sec9x tan9x d/dx(9x)
=9sec9x tan9x
(iv) Let y = sin x² ;
Diff both sides w.r.t x, we have
dy/dx=d/dx(sin x²)=cos x² d/dx(x²)
=2xcos x²
(v) Let y = cos√x ;
Diff both sides w.r.t x, we have
dy/dx=d/dx(cos√x)=-sin√x d/dx(x^1/2)
=-sin√x 1/2x^(-1/2)=-sin√x/2√x
(vi) Let y = 2 cosec bx³ ;
Diff both sides w.r.t A, we have
dy/dx=d/dx(2cosec bx³)=-2cosec bx³ cot bx³ d/dx(bx³)
=-6bcosec bx³ cot bx³
Que-5:
(i) x³/3x-2
(ii)x/sinx
(iii) 1+cosx/1-cosx
(iv) x²-√(1+x)
(v) sin 2x cos² x
(vi) tan4 7x
(vii) 1/sinx + cosx
(viii) 1+cosx/x
Sol:
Que-6: Given that y =sinx-cosx/sinx+cosx show that dy/dx=1+y².
Sol:
Que-7: Differentiate with respect to x :
(i) (2x² – 1) (x³ + 4)³
(ii) x^(4) -1 /√1+x
(iii)(x+1/x)^(-1)
(iv) tan42x
Sol:
Que-8: Find the gradient function dy/dx for each of the following :
(i) y = x – 7x²
(ii) y = 4x7 – 3x³ + 5x – 11.
Sol: (i) Given y = x – 7x² ;
Diff both sides w.r.t x, we have
dy/dx=1-14x
(ii) Given y = 4x7 – 3x³ + 5x – 11
Diff both sides w.r.t x, we have
dy/dx=28x6-9x²+5
Que-9: Find the gradients of the following curves at the points indicated.
(i) y = x² + 5x at (0, 0)
(ii) y = (x + 1) (2x + 3) at (2, 21)
(iii) y = 2x² – x + 4/x at (2, 8)
Sol: (i) Given y = x² + 5x ;
Diff both sides w.r.t x, we have
dy/dx = 2x + 5
at (0, 0); dy/dx =2×0+5=5
(ii) Given y = (x + 1) (2x + 3) = 2x² + 5x + 3
Diff both sides w.r.t x, we have
dy/dx = 4x + 5
∴ at (2, 21); dy/dx = 4×2+5=13
(iii) Given y = 2x² – x + 4/x
Diff both sides w.r.t x, we have
dy/dx=4x-1-4/x²
at (2, 8); dy/dx=4×2-1-4/2²
=8-1-4/4=8-1-1=6
Que-10: If f (x) = 3x² – 4x, find the value of a given that f ‘(a) = 5.
Sol: Given f (x) = 3x² – 4x
Diff both sides w.r.t x, we have f'(x) = 6x – 4
∴ f ‘(a) = 6a – 4
Also f ‘(a) = 5
∴ 5 = 6a – 4 ⇒ 6a = 9 ⇒ a = 3/2
Que-11: Differentiate from first principle.
(i) 3x
(ii) (x + 1) (2x – 3)
(iii) 2-x/4+3x
(iv) x-3/4
(v) √x+1/√x(x>0)
Sol:
–: End Differentiation Class 12 OP Malhotra Exe-8A ISC Math Ch-8 Solution :–
Return to :- OP Malhotra ISC Class-12 S Chand Publication Maths Solutions
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