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

2 Answers
Oct 30, 2017

Answer:

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

Explanation:

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)#

Oct 31, 2017

Answer:

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

Explanation:

#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#