How do you differentiate #e^((2-sqrtx)^2) # using the chain rule?

1 Answer
Nov 5, 2016

Answer:

#y' = e^((2 - sqrt(x))^2) - (2e^((2 - sqrt(x))^2))/sqrt(x)#

Explanation:

This function can be simplified to #e^(4 - 4sqrt(x) + x)#.

We now use the rule that if #y = e^(f(x)),# then #y' = f'(x) xx e^(f(x))#.

So, letting #y = e^square#, with #square = 4 - 4sqrt(x) + x#, we will have to find the derivative of #square#.

#square' = 1 - (0 xx 4sqrt(x) + 1/(2x^(1/2)) xx -4) =1 -2/(x^(1/2))#

So, the derivative is #y' = (1- 2/sqrt(x)) e^((2 -sqrt(x))^2) = e^((2 - sqrt(x))^2) - (2e^((2 - sqrt(x))^2))/sqrt(x)#

Hopefully this helps!