Question 1 :
Differentiate y = (1 + cos2x)6
Solution :
u = 1 + cos2x
Differentiate the function "u" with respect to "x"
du/dx = 0 + 2 cos x (-sin x) ==> -2 sin x cos x
du/dx = - sin 2x
y = u6
Differentiate the function "y" with respect to "x"
dy/dx = 6u5 (du/dx)
= 6(1 + cos2x)5 (-sin 2x)
= -6 sin 2x (1 + cos2x)5
Question 2 :
Differentiate y = e3x/(1 + ex)
Solution :
u = e3x ==> u' = 3e3x
v = 1 + ex ==> v' = ex
dy/dx = [(1 + ex) (3e3x) - e3x (ex)]/(1 + ex)2
dy/dx = [3e3x + 3e4x - e4x]/(1 + ex)2
dy/dx = [3e3x + 2e4x]/(1 + ex)2
Question 3 :
Differentiate y = √(x+√x)
Solution :
y = √(x+√x)
y2 = x + √x
2y (dy/dx) = 1 + 1/2√x
dy/dx = (1/2√(x+√x)) (1 + (1/2√x))
dy/dx = (1/2√(x+√x)) (2√x + 1)/2√x)
dy/dx = (2√x + 1)/4√x(√(x+√x))
Question 4 :
Differentiate y = excos x
Solution :
y = excos x
Let u = x cos x
du/dx = x (-sinx ) + cos x (1)
du/dx = - x sin x + cos x
y = eu
dy/dx = eu (du/dx)
dy/dx = ex cos x (- x sin x + cos x)
Question 5 :
Differentiate the following
Solution :
y2 = x + √(x +√x)
2y (dy/dx) = 1 + 1/2√(x +√x))
Question 6 :
Differentiate y = sin (tan (√sin x))
Solution :
dy/dx = cos (tan (√sin x)) sec2 (√sin x) (1/2√sin x) cos x
dy/dx = [cos (tan (√sin x)) sec2 (√sin x) cos x]/2√sin x
Question 7 :
Differentiate y = sin-1[(1 - x2)/ (1 + x2)]
Solution :
y = sin-1[(1 - x2)/ (1 + x2)]
Let u = (1 - x2)/(1 + x2)
du/dx = [(1 + x2)(-2x) - (1 - x2)(2x)]/(1 + x2)2
= (-2x - 2x3 - 2x + 2x3) /(1 + x2)2
du/dx = -4x /(1 + x2)2
y = sin-1u
dy/dx = 1/√(1 - u2) (du/dx)
= 1/√(1 - ((1 - x2)/(1 + x2))2) [-4x /(1 + x2)2]
= (1 + x2)/√(1 + x4 + 2x2 - 1 + 2x2 - x4)[-4x /(1 + x2)2]
= ((1 + x2)/2x)[-4x /(1 + x2)2]
dy/dx = -2/(1 + x2)
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Nov 02, 24 11:58 PM
Oct 30, 24 10:07 AM
Oct 29, 24 06:24 AM