Question

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

Answer 1

`- (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))`