Differentiation Class 11 OP Malhotra Exe-19D ISC Maths Ch-19 Solutions. In this article you would learn about Derivatives of Trigonometric Functions. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
Differentiation Class 11 OP Malhotra Exe-19D ISC Maths Solutions Ch-19
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-19 | Differentiation |
| Writer | O.P. Malhotra |
| Exe-19(D) | Derivatives of Trigonometric Functions. |
Derivatives of Trigonometric Functions.
Differentiation Class 11 OP Malhotra Exe-19D ISC Maths Ch-19 Solutions.
sin(x): The derivative of sin(x) is cos(x).
cos(x): The derivative of cos(x) is -sin(x).
tan(x): The derivative of tan(x) is sec²(x).
cot(x): The derivative of cot(x) is -csc²(x).
sec(x): The derivative of sec(x) is sec(x)tan(x).
csc(x): The derivative of csc(x) is -csc(x)cot(x).
Que-1: sin 5x
Sol: Let y – sin 5x ; diff. both sides w.r.t. x, we have
dy/dx = (d/dx) sin 5x = cos 5x(d/dx)(5x)
= 5 cos 5x
Que-2: cos 8x
Sol: Let y = cos 8x ; Diff. both sides w.r.t. x, we have
dy/dx = (d/dx) (cos 8x) = -sin8x (d/dx) (8x)
= – 8 sin x
Que-3: sin (5x + 9)
Sol: Let y = sin (5x + 9); Diff. both sides w.r.t. x,
dy/dx = (d/dx) sin (5x + 9) = cos(5x + 9) (d/dx)(5x + 9)
= cos (5x + 9) (5.1 + 0)
= 5 cos (5x + 9)
Que-4: cos (2x – 3)
Sol: Let y = cos (2x – 3) ; diff. both sides w.r.t. x, we have
dy/dx = (d/dx)cos(2x – 3)
= -sin(2x – 3) (d/dx) (2x – 3)
= – 2 sin (2x – 3)
Que-5: tan 7x
Sol: Let y = tan 7x ; Diff. both sides w.r.t. x, we have
dy/dx = (d/dx) tan 7x
= sec² 7x (d/dx) (7x)
= 7 sec² 7x
Que-6: cot nx
Sol: Let y = cot nx ; Diff. both sides w.r.t. x, we get
dy/dx = (d/dx) cot nx
= – cosec² nx (d/dx) (nx)
= – n cosec² nx
Que-7: tan (6x + 11)
Sol: Let y = tan (6x + 11); Diff. both sides w.r.t. x
dy/dx = sec²(6x +11) (d/dx) (6x + 11)
= sec2 (6x + 11) (6.1 + 0)
= 6 sec2 (6x + 11)
Que-8: sin x/3
Sol: Let y = sin (x/3); diff. both sides w.r.t. x, we have
dy/dx = cos (x/3) (d/dx) (x/3)
= 1/3 cos (x/3)
Que-9: sec mx
Sol: Let y = sec mx ; diff. both sides w.r.t. x, we have
dy/dx = (d/dx) (sec mx)
= sec mx tan mx (d/dx) (mx)
= m sec mx tan mx
Que-10: sec [(x/2)-1]
Sol: Let y = sec (x/2−1); Diff. both sides w.r.t. x, we get
dy/dx = (d/dx) sec [(x/2)−1]
= sec [(x/2)−1] tan [(x/2)−1] (d/dx) [(x/2)−1]
= 1/2 sec [(x/2)−1] tan [(x/2)−1]
Que-11: cosec (2/3)x
Sol: Let y = cosec (2/3)x ; Diff. both sides w.r.t. x, we get
dy/dx = – cot (2/3)x cosec (2/3)x (d/dx) (2/3x)
= –2/3 cot (2/3) x cosec (2x/3)
Que-12: x sin x
Sol: Let y = x sin x; Diff. both sides w.r.t. x, we get
dy/dx = (d/dx) (x sin x) = x(d/dx) sin x + sin x (d/dx)(x)
[∵ d/dx (uv) = u(dv/dx) + v (du/dx)]
= x cos x + sin x . 1 = x cos x + sin x
Que-13: x2 cos 5x
Sol: Let y = x² cos 5x; Diff. both sides w.r.t. x, we get
dy/dx = x² (d/dx) cos 5x + cos 5x (d/dx) x²
[∵ d/dx (uv) = u(dv/dx) + v (du/dx)]
= x² (-sin 5x) (d/dx) (5x) + cos 5x . 2x
= – 5x2 sin 5x + 2x cos 5x
Que-14: √x cosec (5x + 7)
Sol: Let y = √x cosec (5x + 7); Diff. both sides w.r.t. x, we have
dy/dx = √x (d/dx)cosec (5x + 7) + cosec (5x + 7) (d/dx) = √x
= √x {- cot (5x + 7) cosec (5x + 7)} (d/dx) (5x + 7) + cosec (5x + 7) (1/2)x^{(1/2)−1}
= -5√x cot (5x + 7) cosec (5x + 7) + (1/2√x) cosec (5x + 7)
Que-15: sin3x/(x−6)
Sol: Let y = sin3x/(x−6); Diff. both sides w.r.t. x, we have
Que-16: cosx/5x
Sol:
Que-17: tanx/(2x+3)
Sol: Let y = tanx/(2x+3); Diff. both sides w.r.t. x, we have
Que-18: {sec(ax−b)}/(x²−2)
Sol: Let y = {sec(ax−b)}/(x²−2); diff. both sides w.r.t. x, we have
Que-19: sin 2x
Sol: Let y = sin 2x; Diff. both sides w.r.t. x, we have
dy/dx = (d/dx) sin 2x = cos 2x (d/dx) (2x)
= 2 cos 2x
Que-20: cos 3x
Sol: Let y = cos 3x; diff. both sides w.r.t. x, we have
dy/dx = (d/dx) (cos 3x) = – sin 3x (d/dx) (3x)
= – 3 sin 3x
Que-21: tan 2x
Sol: Let y = tan 2x; diff. both sides w.r.t. x, we have
dy/dx = (d/dx) (tan 2x) = sec² 2x (d/dx) (2x)
= 2 sec² 2x
Que-22: sin x/2
Sol: Let y = sin (x/2); diff. both sides w.r.t. x, we have
dy/dx = (d/dx) sin (x/2) = cos(x/2) (d/dx) (x/2)
= (1/2) cos (x/2)
Que-23: sec ax
Sol: Let y = sec ax ; diff. both sides w.r.t. x, we get
dy/dx = (d/dx) (sec ax) = sec ax tan ax (d/dx)(ax)
= a sec ax tan ax
Que-24: sec (px + q)
Sol: Let y = sec (px + q); diff. both sides w.r.t. x, we get
dy/dx = (d/dx) sec(px + q)
= sec (px + q) tan (px + q) (d/dx) (px + q)
= p sec (px + q) tan (px + q)
Que-25: tan (4x – 7)
Sol: Let y = tan (4x – 7) ; diff. both sides w.r.t. x,
dy/dx = (d/dx) tan(4x – 7) = sec²(4x – 7) (d/dx)(4x – 7)
= sec² (4x – 7) (4 × 1 – 0)
= 4 sec² (4x – 7)
–: End of Differentiation Class 11 OP Malhotra Exe-19D ISC Maths Ch-19 Solutions. :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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