How do you differentiate #y=5^((3x)/2)#?

1 Answer
Sep 11, 2016

#(3 ln(5)) / (2) cdot 5^((3 x) / (2))#

Explanation:

We have: #y = 5^((3 x) / (2))#

This function can be differentiated using the "chain rule".

Let #u = (3 x) / (2) => u' = (3) / (2)# and #v = 5^(u) => v' = 5^(u) (ln(5))#:

#=> y' = (3) / (2) cdot 5^(u) (ln(5))#

#=> y' = (3 ln(5)) / (2) cdot 5^(u)#

We can now replace #u# with #(3 x) / (2)#:

#=> y' = (3 ln(5)) / (2) cdot 5^((3 x) / (2))#