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Yes. For example, to differentiate y = (x^2 + 1)^x, we take the natural log of both sides.ln(y) = ln((x^2 + 1)^x)

Bring down the exponent.

ln(y) = x ln(x^2 + 1)

Differentiate both sides.

dy/y = ((2x^2)/(x^2 + 1) + ln(x^2 + 1)) dx

Substitute in y = (x^2 + 1)^x.

dy/((x^2 + 1)^x) =((2x^2)/(x^2 + 1) + ln(x^2 + 1)) dx

Solve for dy/dx.

dy/dx = ((x^2 + 1)^x)((2x^2)/(x^2 + 1) + ln(x^2 + 1))

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Q: Is logarithmic differentiation specially useful when dealing with functions with exponent that also depend on x?
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