How do you differentiate #g(x) = (1/(x^3-1))*sqrt(1+e^(x))# using the product rule?

1 Answer
Jan 13, 2016

Answer:

#- (3x^2) / (x^3 - 1)^2 * sqrt(1 + e^x) + 1 / (x^3 - 1) * e^x / (2sqrt(1 + e^x))#

Explanation:

The product rule states: if

#g(x) = f(x) * h(x)#,

the derivative of #g(x)# can be computed as follows:

#g'(x) = f'(x) * h(x) + f(x) * h'(x)#

In your case,

#f(x) = 1 / (x^3 - 1) " "# and

#h(x) = sqrt(1 + e^x)#

The first thing you need to do is differentiate #f(x)# and #h(x)#.

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Let's start with #f(x)#. Here, the chain rule is helpful. You can write use it as follows:

#f(x) = 1/u# where #u = x^3 -1#

According to the chain rule, the derivative is the derivative of #1/u# multiplied with the derivative of #u#.

# [1/u]' = [u^(-1)]' = - u^(-2) = - 1 / u^2 = - 1/(x^3 -1)^2#

# [x^3 -1 ]' = 3x^2#

Thus,

#f'(x) = - 1 / (x^3 - 1)^2 * 3x^2 = - (3x^2) / (x^3 - 1)^2#

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Now, you still need to differentiate #h(x)#. The chain rule can help here as well:

#h(x) = sqrt(v)# where #v= 1 + e^x#

The derivative of #h(x)# is the derivative of #sqrt(v)# multiplied with the derivative of #v#.

#[sqrt(v)]' = [v^(1/2)]' = 1/2 v^(-1/2) = 1 / (2sqrt(v)) = 1 / (2sqrt(1 + e^x))#

# [1 + e^x]' = e^x#

This means that the derivative of #h(x)# is:

#h'(x) = 1 / (2sqrt(1 + e^x)) * e^x = e^x / (2sqrt(1 + e^x))#

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Now, the only thing left to do is applying the product rule:

#g'(x) = f'(x) * h(x) + f(x) * h'(x)#

# color(white)(xxxiii) = - (3x^2) / (x^3 - 1)^2 * sqrt(1 + e^x) + 1 / (x^3 - 1) * e^x / (2sqrt(1 + e^x))#

# color(white)(xxxiii) = - (3x^2 sqrt(1 + e^x) )/ (x^3 - 1)^2 + e^x / ( 2sqrt(1 + e^x)(x^3 - 1))#