1.

Find the derivation of function(3 sec x – 4 cosec x) (2 sin x + 5 cos x)

Answer»

y = (3 sec x – 4 cosec x) (2 sin x + 5cos x) 

Let u = 3 sec x-4 cosec x and v = 2 sin x + 5 cos x 

u’ = 3 (sec x tan x) – 4 (-cosec x cot x) ; v’ = 2 (cos x) + 5 (- sin x) 

u’ = 3 sec x tan x + 4 cosec x cot x); v’ = 2 cos x – 5 sin x. 

∴ y’ = uv’ + vu'

So dy/dx = (3 sec x – 4 cosec x) (2 cos x – 5 sin x) + (2 sin x + 5 cos x) (3 sec x tan x + 4 cosec x cot x) 

= 6 sec x cos x – 15 sec x sin x – 8 cosec x cos x + 20 cosec x sin x + 6 sin x sec x tan x + 8 sin x cosec x cot x + 15 cos x sec x tan x + 20 cos x cosec x cot x 

= 6 (1/cos x) cos x – 15 (1/cos x) sin x – 8 (1/sin x) cos x + 20 (1/sin x) sin x + 6 sin x (1/cos x) tan x + 8 sin x (1/sin x) cot x + 15 cos x (1/cos x) tan x + 20 cos x (1/sin x) cot x 

= 6 – 15 tan x – 8 cot x + 20 + 6 tan2 x + 8 cot x + 15 tan x + 20 cot2 x 

= 26 + 6 tan2 x + 20 cot2 x


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