यदि `e^(x)+e^(y)=e^(x+y)` हो,तो सिद्ध कीजिए कि `(dy)/(dx)=-(e^(x)(e^(y)-1))/(e^(y)(e^(x)-1))`

Explanation

`e^(x)+e^(y)=e^(x+y)`
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
`d/(dx)(e^(x)+e^(y))=d/(dx)(e^(x+y))`
`rArrd/(dx)(e^(x))+d/(dx)(e^(y))=e^(x+y)d/(dx)(x+y)`
`rArre^(x)+e^(y)(dy)/(dx)=e^(x+y)(1+(dy)/(dx))`
`rArre^(x)+(e^(y)-e^(x+y))(dy)/(dx)=e^(x+y)`
`rArr(e^(y)-e^(x+y))(dy)/(dx)=e^(x+y)-e^(x)`
`rArre^(y)(1-e^(x))(dy)/(dx)=e^(x)(e^(y)-1)`
`rArr(dy)/(dx)=(e^(x)(e^(y)-1))/(e^(y)(1-e^(x)))`
`rArr(dy)/(dx)=-(e^(x)(e^(y)-1))/(e^(y)(e^(x)-1))`
यही सिद्ध करना था|