How do you differentiate #g(x) = (x/2)(x-e^(2x))# using the product rule?

1 Answer
Jun 10, 2016

Answer:

#g'(x)=(1-e^(2x))-e^(2x)/2.#

Explanation:

#g'(x)=(x-e^(2x)){d/dx(x/2)}+(x/2){d/dx(x-e^(2x))}=(x-e^(2x))(1/2)+(x/2){d/dx(x)-d/dx(e^(2x))}=(x-e^(2x))(1/2)+(x/2){1-e^(2x)*d/dx(2x)}=(x-e^(2x))(1/2)+(x/2){1-2*e^(2x)}=(1/2){(x-e^(2x))+(x-2x*e^(2x))}=(1/2){2x-(1+2x)*e^(2x)}.#

OR ,
#g'(x)=(1-e^(2x))-e^(2x)/2.#

Note : In fact, as we have to use the product rule, we have done it as above. Otherwise, we can easily derive the solution by first simplifying the function #g(x)# as #(1/2)(x^2 -x*e^(2x))# & then differentiating it!