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Let y = ln(x2), where ln is log to base e.

Using our log laws, y = 2 ln (x)

Therefore dy/dx = 2 . 1/x = 2/x

Another way to do this is to use the chain rule (for differentiation)

Let y = ln(x2), where ln is log to base e.

Let u = x2, therefore du/dx = 2x

y = ln u, therefore dy/du = 1/u

Putting it altogether gives:

dy/dx = dy/du . du/dx = 1/u . 2x = 2x/x2 = 2/x

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Q: How to differentiate ln x 2?
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