1.

यदि `x^(y)+y^(x)=4` , `(dy)/(dx)` निकालें|

Answer» दिया है, `x^(y)+y^(x)=4` … (1)
माना कि `u=x^(y)` तथा `v=y^(x)`
`becauseu=x^(y)`
`thereforelogu=ylogx`
दोनों तरफ x के सापेक्ष अवकलित करने पर हमें मिलता है,
`(1)/(u)(du)/(dx)=y*(1)/(x)+logx*(dy)/(dx)`
`therefore(du)/(dx)=u((y)/(x)+logx(dy)/(dx))=x^(y)((y)/(x)+logx(dy)/(dx))` ...(2)
पुनः `v=y^(x)`
`thereforelogv=xlogy`
दोनों तरफ x के सापेक्ष अवकलित करने पर हमें मिलता है,
`therefore(dv)/(dx)=v(logy+(x)/(y)(dy)/(dx))=y^(x)(logy+(x)/(y)(dy)/(dx))`
(1) से ,u+v=4
`therefore(du)/(dx)+(dv)/(dx)=0`
`rArrx^(y)((y)/(x)+logx(dy)/(dx))+y^(x)*(logy+(x)/(y)(dy)/(dx))=0`
या `(x^(y)logx+y^(x)*(x)/(y))(dy)/(dx)=-(y^(x)logy+(y)/(x)*x^(y))`
या `(x^(y)logx+xy^(x-1))(dy)/(dx)=-(y^(x)logy+yx^(y-1))`
`therefore(dy)/(dx)=-(y^(x)logy+yx^(y-1))/(x^(y)logx+xy^(x-1))`


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