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If y = e1+logx then find dydx

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प्रश्न

If y = `"e"^(1 + logx)` then find `("d"y)/("d"x)` 

योग
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उत्तर

y = `"e"^(1 + logx)` 

= `"e"*"e"^(logx)`

= e. x

∴ `("d"y)/("d"x)` = e. 1 = e

OR

y = `"e"^(1 + logx)` 

`("d"y)/("d"x) = "d"/("d"x)("e"^(1 + logx))`

= `"e"^(1 + logx) * "d"/("d"x)(1 + log x)`

= `"e"^(1 + logx) * (0 + 1/x)`

= `("e"^(1 + log x))/x`

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