# Given f(x)=(e^x)(x-1)/x^2 and g(x)=(x^2)f(x) +1, what is g'(x)?

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It is given that f(x) = (e^x)(x-1)/x^2 and g(x)=(x^2)*f(x) + 1. We have to determine g'(x).

g(x) = (x^2)*f(x) + 1

=> (x^2)(e^x)(x-1)/x^2 + 1

=> (x - 1)e^x + 1

use the product rule to find g'(x)

g'(x) = (x - 1)'e^x + (x - 1)(e^x)'

=> e^x + (x - 1)e^x

=> e^x + x*e^x - e^x

=> x*e^x

**The required derivative g'(x) = x*e^x**

First, we'll determine the function g(x):

g(x) = (x^2)*(e^x)*(x-1)/x^2 + 1

We'll simplify and we'll get:

g(x) = (e^x)*(x-1) + 1

We'll differentiate the function g(x), with respect to x:

g'(x) = [(e^x)*(x-1) + 1]'

We'll apply the product rule:

g'(x) = (e^x)'*(x-1) + (e^x)*(x-1)'

g'(x) = (e^x)*(x-1) + (e^x)

We'll factorize by e^x:

g'(x) = (e^x)*(x-1+1)

We'll eliminate like terms inside brackets:

g'(x) = x*(e^x)

**The first derivative of the function g(x) is g'(x) = x*(e^x).**