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प्रश्न
Differentiate the following w.r.t. x: xe + xx + ex + ee
उत्तर
Let y = xe + xx + ex + ee
Let u = xx
Then log u = logxx = xlogx
Differentiating both sides w.r.t. x, we get
`1/u.(du)/(dx) = d/(dx)(xlogx)`
= `xd/(dx)(logx) + (logx).d/(dx)(x)`
= `x xx (1)/x + (logx)(1)`
∴ `(du)/(dx)` = u(1 + log x) = xx (1 + logx) ...(1)
Now, y = xe + u + ex + ee
∴ `(dy)/(dx) = d/(dx)(x^e) + (du)/(dx) + d/(dx)(e^x) + d/(dx)(e^e)`
= exe–1 + xx (1 + logx) + ex + 0 ...[By (1)]
= exe–1 + xx (1 + logx) + ex
= exe–1 + ex + xx (1 + logx).
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