How do you find the derivative of #y = (ln 2)^x#?

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Answer 1 · 2016-01-19T19:21:18

First, take the natural log of both sides of the equation. Then, use implicit differentiation ...

#y = (ln 2)^x#

#lny = ln[(ln 2)^x]#

Use the property of logs ...

#lny = xln(ln 2)^#

Now, implicit differentiation ...

#(1/y)y'=ln(ln2)#

Finally, solve for #y'#

#y'=yln(ln2)#

#=(ln2)^xln(ln2)#

hope that helped