Tuesday, March 5, 2013

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

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