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Question

Find the differential coefficient of axlogx w.r.t x

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Solution

Let y=axlogx
on differentiation wrt x, we have
dydx=logx×d(ax)dx+ax×d(logx)dx (using chain rule of differentiation)
dydx=logx×ax(ln(a))+ax×1x
dydx=ax(logx ln(a)+1x)
This is the required coefficient.

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