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201.

निम्नलिखित फलनों का अवकलज ज्ञात कीजिए- `sec^(-1)((x^(2)+1)/(x^(2)-1))`

Answer» `y=sec^(-1)((x^(2)+1)/(x^(2)-1))`
माना `x=cottheta`, तब `theta=cot^(-1)x`
`thereforey=sec^(-1)((cot^(2)theta+1)/(cot^(2)theta-1))`
`rArry=sec^(-1)((1+tan^(2)theta)/(1-tan^(2)theta))` ltbr gt`rArry=sec^(-1)((cos^(2)theta+sin^(2)theta)/(cos^(2)theta-sin^(2)theta))`
`rArry=sec^(-1)(1/(cos^(2)theta-sin^(2)theta))`
`rArry=sec^(-1)(1/(cos2theta))`
`y=sec^(-1)(sec2theta)`
`rArry=2theta=2cot^(-1)x`
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
`(dy)/(dx)=2d/(dx)(cot^(-1)x)`
`rArr(dy)/(dx)=2xx(-1)/(1+x^(2))`
`rArr(dy)/(dx)=(-2)/(1+x^(2))`
Discussion
202.

`y=cos^(-1)((1-x^(2))/(1+x^(2)))` जबकि `0ltxlt1` का अवकलज ज्ञात कीजिए

Answer» `y=cos^(-1)((1-x^(2))/(1+x^(2)))`, जहाँ `0ltxlt1`
`x=tantheta` रखने पर,
`thereforey=cos^(-1)((1-tan^(2)theta)/(1+tan^(2)theta))`
`rArry=cos^(-1)(cos2theta)`
`rArry=2theta`
`[because0ltxlt1rArr0lttanthetalt1rArr0ltthetaltpi/4rArr0lt2thetaltpi/2]`
`rArry=2tan^(-1)x,`
`[becausex=tanthetarArrtheta=tan^(-1)x]`
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
`(dy)/(dx)=2d/(dx)tan^(-1)x`
`rArr(dy)/(dx)=2.(1)/(1+x^(2))`
`rArr(dy)/(dx)=2/(1+x^(2))`
Discussion
203.

यदि `y=sin^(-1)[xsqrt(1-x)-sqrtxsqrt(1-x^(2))]` हो,तो `(dy)/(dx)` ज्ञात कीजिए।

Answer» `y=sin^(-1)[xsqrt(1-x)-sqrtxsqrt(1-x^(2))]`
`rArry=sin^(-1)[sqrt(1-(sqrtx)^(2))-sqrtxsqrt(1-x^(2))]`
`x=sintheta` और `sqrtx=sinphi` रखने पर, ltbr gt`y=sin^(-1)[sinthetasqrt(1-sin^(2)phi)-sinphisqrt(1-sin^(2)theta)]`
`rArry=sin^(-1)[sinthetacosphi-sinphicostheta]`
`rArry=sin^(-1)[sin(theta-phi)]`
`rArry=theta-phi`
`rArry=sin^(-1)x-sin^(-1)sqrtx`
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
`(dy)/(dx)=d/(dx)(sin^(-1)x)-d/(dx)(sin^(-1)sqrtx)`
`(du)/(dx)=1/(sqrt(1-x^(2)))-1/(1-(sqrtx)^(2)).d/(dx)(sqrtx)`
`(dy)/(dx)=1/(sqrt(1-x^(2)))-1/(1-x)xx1/(2sqrtx)`
`therefore(dy)/(dx)=1/(sqrt(1-x^(2)))-1/(2sqrtx(1-x))`
Discussion
204.

यदि `x^(m)y^(n)=(x+y)^(m+n)` हो, तो `dy/dx` का मान होगा-A. `(x+y)/(xy)`B. xyC. `(x)/(y)`D. `(y)/(x)`

Answer» Correct Answer - D
Discussion
205.

`sin^(-1)((2x)/(1+x^(2)))` का अवकलज `sin^(-1)((1-x^(2))/(1+x^(2)))` के सापेक्ष है-A. `-1`B. 1C. 2D. `1/2`

Answer» Correct Answer - A
Discussion
206.

यदि `siny=x sin(a+y)` हो, तो `(dy)/(dx)` का मान होगा-A. `(sin(a+y))/(sin a)`B. `(sin^(2)(a+y))/(sin a)`C. `(2sin (a+y))/(sin a)`D. `(sin^(2)(a+y))/(sin y)`

Answer» Correct Answer - B
Discussion
207.

यदि `y=sin^(-1)((2^(x+1))/(1+4^(x)))`, हो,तो `(dy)/(dx)` ज्ञात कीजिए।

Answer» `y=sin^(-1)((2^(x+1))/(1+4^(x)))`
`y=sin^(-1)((2^(x).2)/(1+4^(x)))`
`y=sin^(-1)((2.2^(x))/(1+(2^(x))^(2)))`
माना `2^(x)=tantheta`, तब `theta=tan^(-1)(2^(x))`
`thereforey=sin^(-1)[(2tantheta)/(1+tan^(2)theta)]`
`rArry=sin^(-1)(sin2theta)`
`rArry=2theta=2tan^(-1)2^(x)`
`therefore(dy)/(dx)=d/(dx)(2tan^(-1)2^(x))`
`rArr(dy)/(dx)=2xx1/(1+(2^(x))^(2)).d/(dx)(2^(x))`
`rArr(dy)/(dx)=2/(1+4^(x)).2^(x)log_(e)2`
`rArr(dy)/(dx)=(2^(x+1)(log_(e)2))/(1+4^(x))`
Discussion
208.

`sin^(-1)((1-x^(2))/(1+x^(2)))` का अवकलज ज्ञात कीजिए `0ltxlt1`

Answer» `y=sin^(-1)((1-x^(2))/(1+x^(2)))`, जहाँ `0ltxlt1`.
`x=tantheta` रखने पर,
`y=sin^(-1)((1-tan^(2)theta)/(1+tan^(2)theta))`
`rArry=sin^(-1)(cos2theta)`
`rArry=sin^(-1){sin(pi/2-2theta)}`
`rArry=pi/2-2theta`
`[because0ltxlt1rArr0lttanthetalt1rArr0ltthetaltpi/4rArr0lt2thetaltpi/2rArr0ltpi/2-2thetaltpi/2]`
`rArry=pi/2-2tan^(-1)x`,
`[becausex=tanthetarArrtheta=tan^(-1)x]`
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
`therefore(dy)/(dx)=d/(dx){pi/2-2tan^(-1)x}`
`rArr(dy)/(dx)=0-2(1)/(1+x^(2))`
`rArr(dy)/(dx)=(-2)/(1+x^(2))`
Discussion
209.

`sqrt(sin 3x)` को प्रथम सिद्धांत से अवलंकित कीजिएः

Answer» माना `y=sqrt(sin 3x)`
`Rightarrow y+deltay=sqrt(sin 3(x+deltax))`
`Rightarrow y+deltay-y=sqrt(sin 3(x+8x))-sqrt(sin 3x)`
`Rightarrow (deltay)/(deltax)=(sqrt(sin 3(x+deltax))-sqrt(sin 3x))/(deltax)`
अब `(dy)/(dx)=underset(deltax to 0)lim (deltay)/(deltax)`
`=underset(deltax to 0)lim [({sqrt(sin (3x+3deltax))-sqrt(sin 3x)})/(deltax)xx({sqrt(sin 3x+3deltax)+sqrt(sin3x)})/({sqrt(sin(3x+3deltax))+sqrt(sin 3x)})]`
`=underset(deltax to 0)lim ([sin (3x+3deltax)-sin3x])/(deltax.{sqrt(sin(3x+3deltax))+sqrt(sin 3x)})`
`=underset(deltax to 0)lim (2 cos (3x+(3)/(2)deltax)sin((3)/(2)deltax))/(deltax.{sqrt(sin (3x+3deltax))+sqrt(sin 3x)})`
`=underset(deltax to 0)lim [2cos(3x+(3)/(2)deltax).(sin((3)/(2)deltax))/(((3)/(2)deltax))xx(3)/(2) (1)/(sqrt(sin (3x+3deltax))+sqrt(sin 3x))]`
`=underset(deltax to 0)lim cos (3x+(3)/(2)deltax).underset((3)/(2)deltax to 0)lim (sin((3)/(2)deltax))/(((3)/(2)deltax)).3 underset(deltax to 0)lim (1)/(sqrt(sin(3x+3deltax))+sqrt(sin 3x))`
`=3 cos 3x xx 1 xx (1)/(2sqrt(sin3x))=(3 cos 3x)/(2 sqrtsin 3x)`
Discussion
210.

यदि `x=a[cost+1/2logtan^(2)t/2]` और `y=asint,` तब `(dy)/(dx)` ज्ञात कीजिए।

Answer» यहाँ
`x=a[cost+1/2logtan^(2)t/2]` और y=asint
`rArrx=a{cost+1/2xx2logtant/2}` और y=asint
`rArrx=a{cost+logtant/2}` और y=asint
दोनों पक्षों का t के सापेक्ष अवकलन करने पर,
`(dx)/(dt)=a{-sint+1/(tant/2)(sec^(2)t/2)xx1/2}` और
`(dy)/(dt)=acost`
`rArr(dx)/(dt)=a{-sint+1/(2sin(t/2)cos(t/2))}` और
`(dy)/(dt)=acost`
`rArr(dx)/(dt)=a{-sint+1/(sint)}` और `(dy)/(dt)=acost`
`rArr(dx)/(dt)=a{(-sin^(2)t+1)/(sint)}` और `(dy)/(dt)=acost`
`rArr(dx)/(dt)=(acos^(2)t)/(sint)` और `(dy)/(dt)=acost`
`therefore(dy)/(dx)=(dy//dt)/(dx//dt)`
`rArr(dy)/(dt)=(acost)/((acos^(2)t)/(sint))=tant`
Discussion
211.

`tan^(-1)((acosx=bsinx)/(b cosx+asinx))`

Answer» Correct Answer - -1
`=tan^(-1)(((a)/(b)-tanx)/(1+(a)/(b)tanx))="tan"^(-1)(a)/(b)-tan^(-1)(tanx)="tan"^(-1)(a)/(b)-x`
Discussion