Find the first and second derivatives of the function p = \frac {q^2 + 3}{q - q - q^3}

Question:

Find the first and second derivatives of the function {eq}p = \frac {q^2 + 3}{q - q - q^3} {/eq}

Quotient Rule:

We will apply the quotient rule to the function above to find the derivative.

The second derivative means simply the derivative of the first derivative.

Whenever possible, it is helpful to simplify the function as much as possible before differentiating.

We first simplify the function:

{eq}p= \frac {q^2 + 3}{q - q - q^3} = \frac {q^2 + 3}{ - q^3} {/eq}

Now we differentiate using the quotient rule:

{eq}\begin{align*} p'& = \dfrac{(q^2 + 3)'(-q^3) - (q^2 + 3)(-q^3)'}{(-q^3)^2} \\ & = \dfrac{(2q)(-q^3) - (q^2 + 3)(-3q^2)}{(-q^3)^2} \\ & = \dfrac{-2q^4 - (q^2 + 3)(-3q^2)}{q^6} \\ & = \dfrac{-2q^4 +3 q^4 +9q^2}{q^6} \\ & = \dfrac{q^4 +9q^2}{q^6} \\& = \dfrac{q^2 +9}{q^4} \end{align*} {/eq}

Now we differentiate again for the second derivative using the quotient rule:

{eq}\begin{align*} p'' & = \displaystyle \dfrac{d}{dq}\left (\dfrac{q^2 +9}{q^4} \right) \\& = \dfrac{(q^2 +9)'(q^4) - (q^2 +9)(q^4)'}{(q^4)^2} \\& = \dfrac{(2q)(q^4) - (q^2 +9)(4q^3)}{(q^4)^2} \\& = \dfrac{2q^5 - 4q^5 -36q^3}{q^8} \\& = \dfrac{-2q^5 -36q^3}{q^8} \\& = \dfrac{-2q^2 -36}{q^5} \end{align*} {/eq}