`x` के सापेक्ष निम्नलिखित फलन को अवकलित करें।
`log(logx) ,xgt1`

Options

A. `(1)/(xlogx)`
B. `(1)/(logx)`
C. `(1)/((x^2)logx)`
D. `(2)/(xlogx)`

Answer

A

Explanation

(i) माना कि `y=log(logx),xgt1`
तो `(dy)/(dx)=(d)/(dx)log(logx)=(d)/(dlogx)sin(logx)*(d)/(dx)logx`
`=(1)/(logx)*(1)/(x)=(1)/(xlogx)`
माना कि `y=sin(logx)`
तो `(dy)/(dx)=(d)/(dx)sin(logx)=(d)/(dlogx)sin(logx)*(d)/(dx)logx`
`=cos(logx)*(1)/(x)=(1)/(x)cos(logx)`
(iii) माना कि `y=cos(loge^(x))=cos(xlog_(e)e)=cosx`
इस प्रकार `y=cosx therefore(dy)/(dx)=-sinx`
(iv) माना कि `y=cos(logx+e^(x))`
तो `(dy)/(dx)=(d)/(dx)cos(logx+e^(x))=(d)/(d(logx+e^(x)))cos(logx+e^(x))*(d)/(dx)(logx+e^(x))`
`=-sin(logx+e^(x))*[(d)/(dx)(logx)+(d)/(dx)e^(x)]=-sin(logx+e^(x))((1)/(x)+e^(x))`
माना कि `y=logsinsqrt(x^(2)+1)`.
`therefore(dy)/(dx)=(1)/(sinsqrt(x^(2)+1))*(d)/(dx)(sinsqrt(x^(2)+1))`
`=(1)/(sinsqrt(x^(2)+1))*cossqrt(x^(2)+1)*(d)/(dx)(sqrt(x^(2)+1))`
`=cotsqrt(x^(2)+1)*(1)/(2)(x^(2)+1)^(-1//2)(2x+0)=(xcotsqrt(x^(2)+1))/(sqrt(x^(2)+1))`
(vi) माना कि `y=log_(sinx)cosx`.
तो `y=(logcosx)/(logsinx) " " [ becauselog_(b)a=(loga)/(logb)]`
`therefore(dy)/(dx)=((logsinx)(d)/(dx)(logcosx)-logcosx*(d)/(dx)(logsinx))/((logsinx)^(2))`
`=((logsinx)(1)/(cosx)*(-sinx)-(logcosx)(1)/(sinx)*cosx)/((logsinx)^(2))`
`=-((logsinx)tanx+(logcosx)cotx)/((logsinx)^(2))`
माना कि `y=cos(log(sinx))`
तो `(dy)/(dx)=(d)/(dx)cos(log(sinx))=(d)/(dlogsinx)cos(logsinx)*(d)/(d sinx)logsinx(d)/(dx)sinx`
`=-sin(log(sinx))*(1)/(sinx)*cosx=-cotxsinlog(sinx)`