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प्रश्न
Differentiate the following w.r.t.x: `e^(log[(logx)^2 - logx^2]`
उत्तर
Let y = `e^(log[(logx)^2 − logx^2]`
∴ y = (logx)2 – log x2 ...[∵ elog x = x]
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"[(log x)^2 − logx^2]`
∴ `"dy"/"dx" = "d"/"dx"(log x)^2 − "d"/"dx"(log x^2)`
∴ `"dy"/"dx" = 2 log x. "d"/"dx"(log x) − 1/x^2. "d"/"dx" x^2`
∴ `"dy"/"dx" = 2 log x. 1/x − 1/x^2. 2x`
∴ `"dy"/"dx" = (2log x)/x − 2/x`
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