Differentiate: 1 - cos x 1 + cos x

Given function is 1 - cos x 1 + cos x We know that, from quotient rule u v ' = u ' v - u v ' v 2 , where v ≠ 0 and d x n d x = n x n - 1 . Let us take u = ( 1 - cos x ) and v = ( 1 + cos x ) u ' = d u d x = d ( 1 - cosx ) d x = sin x v ' = d v d x = d ( 1 + cos x ) d x = - sin x Putting the above obtained values in the formula: 1 - cos x 1 + cos x ' = sin x × ( 1 + cos x ) - ( 1 - cos x ) × ( - sin x ) ( 1 + cos x ) 2 = sin x + sin x cos x + sin x - sin x cos x ( 1 + cos x ) 2 = 2 sin x ( 1 + cos x ) 2 Therefore, 2 sin x ( 1 + cos x ) 2

Differentiate: 1 - cos x 1 + cos x

Given function is 1 - cos x 1 + cos x We know that, from quotient rule u v ' = u ' v - u v ' v 2 , where v ≠ 0 and d x n d x = n x n - 1 . Let us take u = ( 1 - cos x ) and v = ( 1 + cos x ) u ' = d u d x = d ( 1 - cosx ) d x = sin x v ' = d v d x = d ( 1 + cos x ) d x = - sin x Putting the above obtained values in the formula: 1 - cos x 1 + cos x ' = sin x × ( 1 + cos x ) - ( 1 - cos x ) × ( - sin x ) ( 1 + cos x ) 2 = sin x + sin x cos x + sin x - sin x cos x ( 1 + cos x ) 2 = 2 sin x ( 1 + cos x ) 2 Therefore, 2 sin x ( 1 + cos x ) 2

Differentiate: sin x - cos x sin x + cos x

Given function is sin x - cos x sin x + cos x We know that, from quotient rule u v ' = u ' v - u v ' v 2 , where v ≠ 0 . Let us take u = ( sin x - cos x ) and v = ( sin x + cos x ) u ' = d u d x = d ( sin x - cos x ) d x = ( cos x + sin x ) v ' = d v d x = d ( sin x + cos x ) d x = ( cos x - sin x ) Putting the above obtained values in the formula: sin x - cos x sin x + cos x ' = ( cos x + sin x ) × ( sin x + cos x ) - ( sin x - cos x ) × ( cos x - sin x ) ( sin x + cos x ) 2 = sin 2 x + cos 2 x + 2 sin x cos x + sin 2 x + cos 2 x - 2 sin x cos x ( sin x + cos x ) 2 = 2 sin 2 x + cos 2 x sin 2 x + cos 2 x + 2 sin x cos x = 2 1 + sin 2 x Therefore, sin x - cos x sin x + cos x ' = 2 ( 1 + sin 2 x )

Differentiate: sec x - tan x sec x + tan x

Given function is sec x - tan x sec x + tan x We know that, from quotient rule u v ' = u ' v - u v ' v 2 , where v ≠ 0 and d x n d x = n x n - 1 . Let us take u = ( sec x - tan x ) and v = ( sec x + tan x ) u ' = du dx = d ( secx - tanx ) dx = secxtanx - sec 2 x v ' = d v d x = d ( sec x + tan x ) d x = sec x tan x + sec 2 x Putting the above obtained values in the formula: u v ' = u ' v - uv ' v 2 sec x - tan x sec x + tan x ' = sec x tan x - sec 2 x ( sec x + tan x ) - ( sec x - tan x ) sec x tan x + sec 2 x ( sec x + tan x ) 2 = sec x tan x - sec 2 x ( sec x + tan x ) - ( sec x - tan x ) ( sec x ) ( tan x + sec x ) ( sec x + tan x ) 2 = ( sec x + tan x ) sec x tan x - sec 2 x - ( sec x - tan x ) ( sec x ) ( sec x + tan x ) 2 = ( sec x + tan x ) sec x tan x - sec 2 x - sec 2 x - sec x tan x ( sec x + tan x ) 2 = ( sec x + tan x ) 2 sec x tan x - 2 sec 2 x ( sec x + tan x ) 2 = 2 sec x [ tan x - sec x ] ( sec x + tan x ) Therefore, sec x - tan x sec x + tan x ' = 2 sec x ( tan x - sec x ) ( sec x + tan x )

Differentiate: e x + sin x 1 + log x

Given function is e x + sin x 1 + log x We know that, from quotient rule u v ' = u ' v - u v ' v 2 , where v ≠ 0 and d x n d x = n x n - 1 . Let us take u = e x + sin x and v = ( 1 + log x ) u ' = d u d x = d e x + sin x d x = e x + cos x v ' = d v d x = d ( 1 + log x ) d x = 1 x Putting the above obtained values in the formula: u v ' = u ' v - uv ' v 2 e x + sin x 1 + log x ' = e x + cos x × ( 1 + log x ) - e x + sin x × 1 x ( 1 + log x ) 2 = x e x + cos x ( 1 + log x ) - e x + sin x x ( 1 + log x ) 2 Therefore, e x + sin x 1 + log x ' = x e x + cos x ( 1 + log x ) - e x + sin x x ( 1 + log x ) 2

Differentiate: e x sin x sec x

Given function is e x sinx secx We know that, from quotient rule u v ' = u ' v - u v ' v 2 , where v ≠ 0 , ( f g ) ' = f ' g + f g ' . Let us take u = e x sinx and v = ( sec x ) u ' = d u d x = d e x sin x d x Applying Product rule ( f g ) ' = f ' g + f g ' , Taking g = e x and h = sin x = e x sin x + e x cos x u ' = e x sin x + e x cos x v ' = d v d x = d ( sec x ) d x = sec x tan x Putting the above obtained values in the formula: u v ' = u ' v - uv ' v 2 e x sin x sec x ' = e x sin x + e x cos x × ( sec x ) - e x sin x × ( sec x tan x ) ( sec x ) 2 = e x sin x + e x cos x - e x sin x × ( tan x ) ( sec x ) = cos x e x sin x + e x cos x - e x sin x × ( tan x ) = e x sin x cos x + e x cos 2 x - e x sin x cos x × ( tan x ) = e x sin x cos x + e x cos 2 x - e x sin 2 x = e x sin x cos x + e x cos 2 x - e x sin 2 x = e x sin x cos x + e x cos 2 x = e x ( sin x cos x + cos 2 x ) Therefore, e x sin x sec x ' = e x ( sin x cos x + cos 2 x )

Differentiate: 2 x cot x x

Given function is 2 x cot x x We know that, from quotient rule u v ' = u ' v - u v ' v 2 , where v ≠ 0 , ( g h ) ' = g ' h + g h ' and d x n d x = n x n - 1 Let us take u = 2 x cot x and v = ( x ) u ' = du dx = d 2 x cotx dx Applying Product rule ( g h ) ' = g ' h + g h ' , Taking g = 2 x and h = cot x = 2 x log 2 cot x + 2 x - cosec 2 x u ' = 2 x log 2 cot x - 2 x cosec 2 x u ' = 2 x log 2 cot x - cosec 2 x v ' = d v d x = d ( x ) d x = 1 2 x Putting the above obtained values in the formula: u v ' = u ' v - uv ' v 2 2 x cot x x ' = 2 x log 2 cot x - cosec 2 x × x - 2 x cot x × 1 2 x ( x ) 2 = 2 x log 2 cot x - cosec 2 x × x - 2 x cot x × 1 2 x x = 2 x log 2 cot x - cosec 2 x × x - 2 x - 1 cot x × 1 x x = x 2 x log 2 cot x - cosec 2 x - 2 x - 1 cot x x x = 2 x x log 2 cot x - x cosec 2 x - 2 x - 1 cot x x 3 2 Therefore, 2 x cot x x ' = 2 x - x cosec 2 x + x ( log 2 ) cot x - 1 2 log x x 3 / 2

Differentiate: e x ( x - 1 ) ( x + 1 )

Given function is e x ( x - 1 ) ( x + 1 ) We know that, from quotient rule u v ' = u ' v - u v ' v 2 , where v ≠ 0 , ( g h ) ' = g ' h + g h ' and d x n d x = n x n - 1 Let us take u = e x ( x - 1 ) and v = ( x + 1 ) u ' = d u d x = d e x ( x - 1 ) d x Applying Product rule ( g h ) ' = g ' h + g h ' , Taking g = e x and h = x - 1 e x ( x - 1 ) ' = e x ( x - 1 ) + e x ( 1 ) = e x ( x - 1 ) + e x u ' = e x x v ' = d v d x = d ( x + 1 ) d x = 1 Putting the above obtained values in the formula: u v ' = u ' v - uv ' v 2 e x ( x - 1 ) ( x + 1 ) ' = e x x ( x + 1 ) - e x ( x - 1 ) ( 1 ) ( x + 1 ) 2 = e x x 2 + e x x - e x x + e x ( x + 1 ) 2 = e x x 2 + e x ( x + 1 ) 2 = e x x 2 + 1 ( x + 1 ) 2 Therefore, e x ( x - 1 ) ( x + 1 ) ' = e x x 2 + 1 ( x + 1 ) 2

Differentiate: x tan x ( sec x + tan x )

Given function is x tan x ( sec x + tan x ) We know that, from quotient rule u v ' = u ' v - u v ' v 2 , where v ≠ 0 , ( f g ) ' = f ' g + f g ' and d x n d x = n x n - 1 Let us take u = ( x tan x ) and v = ( sec x + tan x ) u ' = d u d x = d [ x tan x ] d x Applying Product rule ( f g ) ' = f ' g + f g ' , Taking g = x and h = tan x [ x tan x ] ' = ( 1 ) ( tan x ) + x sec 2 x = tan x + x sec 2 x u ' = tan x + x sec 2 x v ' = d v d x = d ( sec x + tan x ) d x = sec x tan x + sec 2 x v ' = sec x ( tan x + sec x ) Putting the above obtained values in the formula: u v ' = u ' v - uv ' v 2 x tan x ( sec x + tan x ) ' = tan x + x sec 2 x ( sec x + tan x ) - [ x tan x ] [ sec x ( tan x + sec x ) ] ( sec x + tan x ) 2 = ( sec x + tan x ) tan x + x sec 2 x - ( x tan x ) ( sec x ) ( sec x + tan x ) 2 = tan x + x sec 2 x - x tan x sec x ( sec x + tan x ) Therefore, x tan x ( sec x + tan x ) ' = x sec x ( sec x - tan x ) + tan x ( sec x + tan x )

Differentiate: a x 2 + b x + c p x 2 + q x + r

Given function is a x 2 + b x + c p x 2 + q x + r We know that, from quotient rule u v ' = u ' v - u v ' v 2 , where v ≠ 0 and d x n d x = n x n - 1 . Let us take u = a x 2 + b x + c and v = p x 2 + q x + r u ' = d u d x = d a x 2 + b x + c d x = 2 a x + b v ' = d v d x = d p x 2 + q x + r d x = 2 p x + q Putting the above obtained values in the formula: u v ' = u ' v - uv ' v 2 a x 2 + b x + c p x 2 + q x + r ' = ( 2 a x + b ) p x 2 + q x + r - a x 2 + b x + c [ 2 p x + q ] p x 2 + q x + r 2 = 2 a p x 3 + 2 a q x 2 + 2 a x r + b p x 2 + b q x + b r - 2 a p x 3 + q a x 2 + 2 b p x 2 + b q x + 2 p c x + c q p x 2 + q x + r 2 = ( a q - b p ) x 2 + 2 ( r a - p c ) x + b r - c p p x 2 + q x + r 2 Therefore, a x 2 + b x + c p x 2 + q x + r ' = ( a q - b p ) x 2 + 2 ( a r - p c ) x + ( b r - c q ) p x 2 + q x + r 2