derivative of x*loge(x) = loge(x) + 1?

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answers:
MyRank say: x log ex = loge x + 1

Differentiation with respect to ‘x’

(1) loge x+ x1/x = 1/x + 0

loge x + 1 = 1/x

loge x - 1/x + 1 = 0

x loge x+ x - 1 = 0
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Roger say: Yes
Use the chain rule
d(uv)/dx = u(dv/dx) + v(du/dx)
with u = x and v = ln(x) since loge(x) = ln(x)
du/dx =1 and dv/dx = 1/x
so u(dv/dx) = x(1/x) = 1
and v(du/dx) = ln(x)
1 + ln(x)
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Φ² = Φ+1 say: The derivative of x ln(x) is 1 × ln(x) + x × 1/x which simplifies to ln(x) + 1, so yes.
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JIM say: The anwser iz big chungus
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