How do you differentiate #f(x)=e^(-2x^2+x+8) # using the chain rule?

1 Answer
Mar 17, 2016

Answer:

#f'(x)=(1-4x)e^(-2x^2+x+8)#

Explanation:

The chain rule states that derivative of a function of a function, say

#d/(dx)f(g(x))=(df)/(dg) * (dg)/(dx)#

So if #f(x)=e^(-2x^2+x+8)#, (here #f(x)=e^g(x)#,

where #g(x)=(-2x^2+x+8)#. Hence

#(df)/(dx)=(de^(-2x^2+x+8))/(d(-2x^2+x+8))*d/dx(-2x^2+x+8)#

#(df)/(dx)=e^(-2x^2+x+8)*d/dx(-2x^2+x+8)#

= #e^(-2x^2+x+8)*(-4x+1)#

= #(1-4x)e^(-2x^2+x+8)#