Consider the function $q(p) = \dfrac{200}{p+4}$. Determine $p'(4)$.
We calculate this derivative in two ways: by the use of the Derivative inverse function and by the use of $p(q)$. We first use the new rule. We know that
$$p'(q) = \dfrac{1}{q'(p(q))} \qquad \text{hence} \qquad p'(4) = \dfrac{1}{q'(p(4))}.$$
In order to use this, we first have to determine $q'(p)$ and $p(4)$. The derivative of $q(p) = 200(p+4)^{-1}$ is
$$q'(p) = 200 \cdot (-1) (p+4)^{-1-1} \cdot 1 = -200(p+4)^{-2} = \dfrac{-200}{(p+4)^2}.$$
Note that $p(4)$ is nothing else than the value of $p$ such that $q$ is equal to 4, which means that we have to solve $q(p) = 4$:
$$
\begin{align}
4 &= \dfrac{200}{p+4}\\
4(p+4) &= 200\\
p+4 &= \dfrac{200}{4} = 50\\
p &= 50-4 = 46.
\end{align}
$$
Hence, $p(4) = 46$.
Now we can determine $p'(4)$:
$$p'(4) = \dfrac{1}{q'(p(4))} = \dfrac{1}{q'(46)} = \dfrac{1}{\tfrac{-200}{(46+4)^2}}= \dfrac{1}{\tfrac{-200}{2500}} = \dfrac{2500}{-200} = -12\tfrac{1}{2}.$$
In this particular case we have the prescription $p(q) = \dfrac{200}{q} - 4 = 200q^{-1} - 4$ (see Inverse function: example 1) and hence, we can use it directly. We obtain
$$
\begin{align}
p'(q) &= 200 \cdot (-1) q^{-1-1} + 0 = -200q^{-2} = \dfrac{-200}{q^2}\\
p'(4) &= \dfrac{-200}{4^2} = \dfrac{-200}{16} = -12\tfrac{1}{2}.
\end{align}
$$
Both methods result in the same answer, but the second is clearly shorter. Unfortunately, we cannot always use it, as we need the prescription of the inverse function, which cannot always be obtained (see Example 2).
We calculate this derivative in two ways: by the use of the Derivative inverse function and by the use of $p(q)$. We first use the new rule. We know that
$$p'(q) = \dfrac{1}{q'(p(q))} \qquad \text{hence} \qquad p'(4) = \dfrac{1}{q'(p(4))}.$$
In order to use this, we first have to determine $q'(p)$ and $p(4)$. The derivative of $q(p) = 200(p+4)^{-1}$ is
$$q'(p) = 200 \cdot (-1) (p+4)^{-1-1} \cdot 1 = -200(p+4)^{-2} = \dfrac{-200}{(p+4)^2}.$$
Note that $p(4)$ is nothing else than the value of $p$ such that $q$ is equal to 4, which means that we have to solve $q(p) = 4$:
$$
\begin{align}
4 &= \dfrac{200}{p+4}\\
4(p+4) &= 200\\
p+4 &= \dfrac{200}{4} = 50\\
p &= 50-4 = 46.
\end{align}
$$
Hence, $p(4) = 46$.
Now we can determine $p'(4)$:
$$p'(4) = \dfrac{1}{q'(p(4))} = \dfrac{1}{q'(46)} = \dfrac{1}{\tfrac{-200}{(46+4)^2}}= \dfrac{1}{\tfrac{-200}{2500}} = \dfrac{2500}{-200} = -12\tfrac{1}{2}.$$
In this particular case we have the prescription $p(q) = \dfrac{200}{q} - 4 = 200q^{-1} - 4$ (see Inverse function: example 1) and hence, we can use it directly. We obtain
$$
\begin{align}
p'(q) &= 200 \cdot (-1) q^{-1-1} + 0 = -200q^{-2} = \dfrac{-200}{q^2}\\
p'(4) &= \dfrac{-200}{4^2} = \dfrac{-200}{16} = -12\tfrac{1}{2}.
\end{align}
$$
Both methods result in the same answer, but the second is clearly shorter. Unfortunately, we cannot always use it, as we need the prescription of the inverse function, which cannot always be obtained (see Example 2).