x के सापेक्ष `y=sqrt(3x+2)+1/sqrt(2x^(2)+4)+(cos x)^("tan x")` का अवलन कीजिये ।

Explanation

`y=sqrt(3x+2)+(1)/(sqrt(2x^(2)+4))+(cos x)^( tan x)`
`rArr y = u + v + w ` ...(1)
जहाँ `u=sqrt(3x+2)` ...(2)
`v=(1)/(sqrt(2x^(2)+4))`
और `:. (dy)/(dx)=(du)/(dx)+(dv)/(dx)+(dw)/(dx)` ...(4)
`u=sqrt(3x+2)`
`:. (du)/(dx)=(1)/(2)(3x+2)^(1/2-1)(d)/(dx) (3x+2)`
`rArr (3)/(2sqrt(3x+2))` ...(5)
`v=(1)/(sqrt(2x^(2)+4))=(2x^(2)+u)^(-1/2)`
`:. (dv)/(dx)=-(1)/(2)(2x^(2)+4)^(-1/2-1)(d)/(dx)(2x^(2)+4)`
`=-(1)/(2(2x^(2)+4)^(3//2)).(4x)`
`=-(2x)/((2x^(2)+4)^(3//2))` ...(6)
`w=(cos x )^(tan x)`
`rArr log w = tan x log cos x`
`rArr (1)/(w) (dw)/(dx)=sec^(2)xlog cos x +tan x . (1)/(cos x)(-sin x)`
`=sec^(2)xlog cos x - tan^(2)x`
`rArr (dw)/(dx)=w(sec^(2)xlog cos x - tan ^(2)x)`
`rArr (dw)/(dx)=(cos x )^(tan x) {sec^(2)x log cos x - tan^(2) x}` ...(7)
समीकरण (4), (5) (6) और (7) से,
`(dy)/(dx)=(3)/(2sqrt(3x+2))-(2x)/((2x^(2)+4)^(3//2))+ (cos x )^(tan x ) { sec^(2) x log cos x - tan^(2)x}`