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