Consider the function $y(x) = x^5 + 5x + 3$. Determine $x'(3)$.
In this case there is no prescription of $x(y)$, which implies that we have to determine $x'(3)$ by the use of the Derivative inverse function. We know that
$$x'(y) = \dfrac{1}{y'(x(y))} \qquad \text{hence} \qquad x'(3) = \dfrac{1}{y'(x(3))}.$$
In order to use this we first need to determine $y'(x)$ and $x(3)$. The derivative of $y(x) = x^5 + 5x + 3$ is
$$y'(x) = 5x^4 + 5.$$
Note that $x(3)$ is the value of $x$ such that $y$ is equal to 3, which means that we have to solve $y(x) = 3$:
$$
\begin{align}
3 &= x^5 + 5x + 3\\
0 &= x^5 + 5x = x(x^4 + 5)\\
x = 0 &\mbox{ or } x^4 + 5 = 0\\
&\phantom{\vee} ~~~ x^4 = -5 \qquad \text{impossible}
\end{align}
$$
Since $x^4$ is non-negative $x^4 = -5$ does not provide a solution. The unique solution is therefore $x(3)=0$.
Now we can determine $x'(3)$:
$$x'(3) = \dfrac{1}{y'(x(3))} = \dfrac{1}{y'(0)} = \dfrac{1}{5\cdot0^4 + 5} = \dfrac{1}{5}.$$
Hence, it is possible to determine the derivative of $x(y)$ without knowing the function $x(y)$.
