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Question
Differentiate the following w.r.t. x: (sin xx)
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Solution
Let y = (sin xx)
Then `"dy"/"dx" = "d"/"dx"[(sinx^x)]`
∴ `"dy"/"dx" = cos(x^x)."d"/"dx"(x^x)` ...(1)
Let u = xx
Then log u = logxx = x.logx
Differentiating both sides w.r.t. x, we get
`1/u."du"/"dx" = "d"/"dx"(x.logx)`
= `x."d"/"dx"(logx) + (logx)."d"/"dx"(x)`
= `x xx (1)/x + (logx) xx 1`
∴ `"du"/"dx" = u(1 + logx)`
∴ `"d"/"dx"(x^x) = x^x (1 + logx)` ...(2)
From (1) and (2), we get
`"dy"/"dx" = cos(x^x).x^x(1 + logx)`
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