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यदि (If) `y=a^(u)`, सिद्ध करें कि (prove that) `(dy)/(dx)=a^(u)loga(du)/(dx)`.

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दिया है, `y=a^(u) thereforelogy=u loga`
दोनों तरफ x के सापेक्ष अवकलित करने पर हमें मिलता है,
`(1)/(y)(dy)/(dx)=(loga)(du)/(dx) " " therefore(dy)/(dx)=yloga(du)/(dx)`
अतः `(dy)/(dx)=a^(u)loga(du)/(dx) " " [becausey=a^(u)]`

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