# Find second derivative

## Homework Statement

f(x) = √x^2 + x + 1

Chain Rule

## The Attempt at a Solution

f ' (x) = (x^2+x+1)^1/2
= 1/2 (x^2+x+1)^-1/2(x^2+x+1)'
= 1/2(x^2+x+1)^-1/2(2x + 1)
= 1/2(2x+1)/√(x^2+x+1)

f '' (x) = So I am having trouble with that. Unless my answer for the first derivative is incorrect. I've already tried the quotient rule as well as the product rule. I haven't succeed in any.

Here, I'll use the product rule:

f '' (x) = 1/2(2x+1)(x^2+x+1)^-1/2

= (x+1/2)' (x^2+x+1)^-1/2 + (x+1/2)(x^2+x+1)^-1/2'

= (1)(x^2+x+1)^-1/2 + (x+1/2)(-1/2)(x^2+x+1)^-3/2(x^2+x+1)'

= (x^2+x+1)^-1/2 + (-1/2x - 1/4)(x^2+x+1)^-3/2(2x+1)

= (x^2+x+1)^-1/2 + (-1 -1/2x - 1/2x - 1/4)(x^2+x+1)^-3/2

= (x^2+x+1)^-1/2 + (-4-4x-1/4)(x^2+x+1)^-3/2

= (-5/4 - x)(x^2+x+1)^-3/2 + 1/(x^2+x+1)^1/2

I know for sure this is not the right answer.

The right answer should be 3/4(x^2+x+1)^3/2

## Answers and Replies

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Mark44
Mentor

## Homework Statement

f(x) = √x^2 + x + 1
Based on your work below, you need parentheses. I think this is your function -
f(x) = √(x2 + x + 1)

Chain Rule

## The Attempt at a Solution

f ' (x) = (x^2+x+1)^1/2
No, the right side isn't the original function. You should indicated that the right side is the derivative of what you started with. Also, although you have some parentheses now, you need some around the exponents

f ' (x) = d/dx [(x^2+x+1)^(1/2)]
= 1/2 (x^2+x+1)^-1/2(x^2+x+1)'
= 1/2(x^2+x+1)^-1/2(2x + 1)
OK except for missing parens around the exponent.
= 1/2(2x+1)/√(x^2+x+1)

f '' (x) = So I am having trouble with that. Unless my answer for the first derivative is incorrect. I've already tried the quotient rule as well as the product rule. I haven't succeed in any.

Here, I'll use the product rule:

f '' (x) = d/dx 1/2(2x+1)(x^2+x+1)^-1/2
It would have been easier to just pull out that factor of 1/2, rather than working with it as you did below.
= (x+1/2)' (x^2+x+1)^-1/2 + (x+1/2)(x^2+x+1)^-1/2'

= (1)(x^2+x+1)^-1/2 + (x+1/2)(-1/2)(x^2+x+1)^-3/2(x^2+x+1)'

= (x^2+x+1)^-1/2 + (-1/2x - 1/4)(x^2+x+1)^-3/2(2x+1)

= (x^2+x+1)^-1/2 + (-1 -1/2x - 1/2x - 1/4)(x^2+x+1)^-3/2

= (x^2+x+1)^-1/2 + (-4-4x-1/4)(x^2+x+1)^-3/2

= (-5/4 - x)(x^2+x+1)^-3/2 + 1/(x^2+x+1)^1/2

I know for sure this is not the right answer.

The right answer should be 3/4(x^2+x+1)^3/2
Your first derivative is correct. Your work for the 2nd derivative might be correct, but the way you did it makes it difficult to check.

Start with f'(x) = (1/2)(2x + 1)(x2 + x + 1)^(-1/2)

f''(x) = (1/2)[ (2x + 1)(-1/2)(x2 + x + 1)^(-3/2)(2x + 1) + 2(x2 + x + 1)^(-1/2)]

Do NOT bring that first factor of 1/2 inside the brackets!

If you clean the above up, by factoring (x2 + x + 1)-3/2 from the two factors inside the brackets, you should end up with the correct answer.

Last edited:
Thank you so much for the reply. Before, I'd used the quotient rule. I see that how you put it, f''(x) = (1/2)[ (2x + 1)(-1/2)(x2 + x + 1)^(-3/2)(2x + 1), is a much better way to rewrite it.

However, I am confused with the second part 2(x2] + x + 1)^(1/2)]. Where is this coming from?

I am so sorry...this is a new topic in my class and I am still trying to understand it.

Thanks again!

Mark44
Mentor
Thank you so much for the reply. Before, I'd used the quotient rule. I see that how you put it, f''(x) = (1/2)[ (2x + 1)(-1/2)(x2 + x + 1)^(-3/2)(2x + 1), is a much better way to rewrite it.

However, I am confused with the second part 2(x2] + x + 1)^(1/2)]. Where is this coming from?
Product rule. Also, I notice that I had an extra bracket that shouldn't have been there, and I forgot to put in the sign for the exponent. I have edited my previous post, so it should be OK now.

What you have above should be
2(x2 + x + 1)^(-1/2)

The first 2 is there because it is the derivative of (2x + 1).

The problem was to find the derivative of (1/2)(2x + 1)(x2 + x + 1)^(-1/2).

f''(x) = (1/2) * d/dx[(2x + 1)(x2 + x + 1)^(-1/2)]
= (1/2) * [ (2x + 1) * d/dx(x2 + x + 1)^(-1/2)) + d/dx(2x + 1) * (x2 + x + 1)^(-1/2)]
I am so sorry...this is a new topic in my class and I am still trying to understand it.

Thanks again!