# Calculating the second derivative.

• Nov 10th 2013, 12:53 PM
sepoto
Calculating the second derivative.
$\displaystyle \frac{x}{x+5}$

I have an answer in my solutions manual that I think might be a typo. It is my understanding right now that the second derivative of x is 0. So if that is correct then the second derivative of the equation at the top of my post should then be:

$\displaystyle \frac{0}{0}$

I want to make sure I have not made an error.

$\displaystyle 0$

• Nov 10th 2013, 01:08 PM
SlipEternal
Re: Calculating the second derivative.
Quote:

Originally Posted by sepoto
$\displaystyle \frac{x}{x+5}$

I have an answer in my solutions manual that I think might be a typo. It is my understanding right now that the second derivative of x is 0. So if that is correct then the second derivative of the equation at the top of my post should then be:

$\displaystyle \frac{0}{0}$

I want to make sure I have not made an error.

$\displaystyle 0$

Yes, you made an error. You can apply the quotient rule. You can apply the product rule to $\displaystyle x(x+5)^{-1}$. But, you cannot take the derivative of the numerator, then the derivative of the denominator, and assume you are getting a correct derivative.

You can also simplify the expression: $\displaystyle \dfrac{x}{x+5} = \dfrac{x+(5-5)}{x+5} = \dfrac{x+5}{x+5} - \dfrac{5}{x+5} = 1-\dfrac{5}{x+5} = 1-5(x+5)^{-1}$

Then, the derivative would be $\displaystyle 0-5(-1)(x+5)^{-2} = 5(x+5)^{-2}$ and the second derivative would be $\displaystyle -10(x+5)^{-3}$.
• Nov 10th 2013, 01:15 PM
SlipEternal
Re: Calculating the second derivative.
Note that I am assuming you know the chain rule. If you do not know the chain rule, then you cannot take the derivative of $\displaystyle (x+5)^{-1}$. Instead, you would need to use the quotient rule:

$\displaystyle \dfrac{d}{dx}\dfrac{x}{x+5} = \dfrac{(x+5)\tfrac{d}{dx}(x) - x\tfrac{d}{dx}(x+5)}{(x+5)^2} = \dfrac{(x+5)\cdot 1 - x\cdot 1}{(x+5)^2} = \dfrac{5}{x^2+10x+25}$

Then apply the quotient rule again to find the second derivative.