InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
निम्नलिखित फलन को x के सापेक्ष अवकलित करें । `sin^(-1)(cosx)` |
|
Answer» (i) माना कि `y=sin^(-1)(cosx)` u= cosx रखें ताकि `y=sin^(-1)u` `therefore(du)/(dx)=-sinx` तथा `(dy)/(du)=(1)/(sqrt(1-u^(2)))` अब `(dy)/(dx)=(dy)/(du)*(du)/(dx)` [Chain rule से] `=(1)/(sqrt(1-u^(2)))*(-sinx)=(-sinx)/(sqrt(-cos^(2)x))=(-sinx)/(sinx)=-1` Second method: `(dy)/(dx)=(d)/(dx)"sin"^(-1)(cosx)=(d)/(d cosx) sin^(-1) (cosx)*(d)/(dx)(cosx)` `=(1)/(sqrt(1-cos^(2)x))*(-sinx)=-(sinx)/(sinx)=1` (ii) माना कि `y=cot^(-1)sqrt(x)` `(dy)/(dx)=(d)/(dx)(cot^(-1)sqrt(x))(d)/(dsqrt(x))cot^(-1)sqrt(x)*(d)/(dx)sqrt(x)` `=-(1)/(1+(sqrt(x))^(2))*(1)/(2sqrt(x))=-(1)/(2sqrt(x)(1+x))` (iii) माना कि `y=sin(2sin^(-1)x)` अब `(dy)/(dx)=(d sin(2sin^(-1)x))/(dx)=(d)/(d(2sin^(-1)x))sin(2sin^(-1)x)*(d)/(dx)(2sin^(-1)x)` `=cos(2sin^(-1)x)*2*(1)/(sqrt(1-x^(2)))=(2)/(sqrt(1-x^(2)))cos(2sin^(-1)x)` (iv) माना कि `y=tan^(-1)(secx+tanx)` `(dy)/(dx)=(d)/(dx)tan^(-1)(secx+tanx)` `=(d)/(d(secx+tanx))tan^(-1)(secx+tanx)*(d)/(dx)(secx+tanx)` `=(1)/((1+sec^(2)x+tan^(2)x+2secxtanx))(secxtanx+sec^(2)x)` `=(secxtanx+sec^(2)x)/((2sec^(2)x+2secxtanx))=(secx(tanx+secx))/(2secx(secx+tanx))=(1)/(2)` `(dy)/(dx)=(d)/(dx){cos(sin^(-1)x)}` `=-(sin(sin^(-1)x))/(sqrt(1-x^(2)))` |
|