derivative of 2xx Solve using logarithmic differentiation So

Solution

y = (2x)^x

ln y = ln (2x)^x

ln y = x ln 2x

ln y = x (ln 2 + ln x)

ln y = x ln 2 + x ln x

d/dx (ln y) = d/dx (x ln 2 + x ln x)

y\'/y = ln 2 + ln x + x/x

y\'/y = ln 2 + ln x + 1

y\'/y = ln x + ln 2 + 1

y\' = y(ln x + ln 2 + 1)

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

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