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

1 Answer
Jan 19, 2016

Answer:

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

Explanation:

#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