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

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Answer 1 · 2016-09-11T12:43:30

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

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))#