How do you differentiate #5^(2x^2)#?

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Answer 1 · 2017-10-30T16:41:52

#d/dx (5^(2x^2)) = 4ln5 x 5^(2x^2)#

Consider that:

#5^(2x^2) = (e^ln5)^(2x^2) = e^(2ln5x^2)#

So, using the chain rule:

#d/dx (5^(2x^2)) = d/dx(e^(2ln5x^2)) = e^(2ln5x^2) d/dx (2ln5x^2)#

#d/dx (5^(2x^2)) = d/dx(e^(2ln5x^2)) = 4ln5xe^(2ln5x^2) = 4ln5 x 5^(2x^2)#

Answer 2 · 2017-10-31T08:35:30

#(dy)/(dx)=5^(2x^2)4xln5#

#y=5^(2x^2)#

take natural logs of both sides

#lny=ln5^(2x^2)#

using laws of logs

#lny=2x^2ln5#

differentiate implicitly

#1/y(dy)/(dx)=4xln5#

#(dy)/(dx)=y4xln5#

#:.(dy)/(dx)=5^(2x^2)4xln5#