Asked by Casey
derivative of y=ln2^x
please help, I am getting the wrong answer every time
please help, I am getting the wrong answer every time
Answers
Answered by
Steve
not sure what you have there
ln(2^x)?
ln(2^x) = x*ln2
y' = ln2
y = log<sub>2</sub>x ?
y = lnx/ln2
y' = 1/(x*ln2)
ln(2^x)?
ln(2^x) = x*ln2
y' = ln2
y = log<sub>2</sub>x ?
y = lnx/ln2
y' = 1/(x*ln2)
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