### Implicit differentiation

When a function is not given explicitly in terms of the independant variable $x$, but rather implicitly in an equation involving the independant variable $x$ and the dependant variable $y$, then we find the derivative by a technique popularly called implicit differentiation method.

As a matter of fact, we have already seen this technique employed when we found the derivative of an inverse function. Given $y=f\left(x\right)$ and $x=g\left(y\right)$ , we have $x=g\left(f\left(x\right)\right)$. Then we differentiate both sides with respect to $x$ to get $1={g}^{\prime }\left(y\right){f}^{\prime }\left(x\right)={g}^{\prime }\left(y\right)\frac{dy}{dx}$.

In this, we have a function $g\left(y\right)$ where $y$ is a function of $x$ , but is not given explicitly, and we are saying that $\frac{d}{dx}\left(g\left(y\right)\right)={g}^{\prime }\left(y\right)\frac{dy}{dx}$.

So, when we have an equation in $x$ and $y$, which defines $y$ as a function of $x$ implicitly, then we differentiate both sides with respect to $x$, and when doing so we employ the technique that $\frac{d}{dx}\left(g\left(y\right)\right)={g}^{\prime }\left(y\right)\frac{dy}{dx}$.

Let us assume that $x={y}^{2}$ defines $y$ as function of $x$ under certain restriction. If we want to find the derivative of $y$ with respect to $x$, we can try to find the function which should not be too difficult in this case and then find the derivative, or we can use the implicit differentiation technique to find it.

Let us differentiate both sides of this equation

$x={y}^{2}$

with respect to $x$ and get

$1=2y\frac{dy}{dx}$

When differentiating ${y}^{2}$ with respect to $x$ , we employed the chain rule $\frac{d}{dx}\left(g\left(y\right)\right)={g}^{\prime }\left(y\right)\frac{dy}{dx}$

Now, diving both sides by $2y$ , we get $\frac{dy}{dx}=\frac{1}{2y}$

There is no need to express the derivative explicitly in terms of the independant varibale $x$. In most cases, it will simply be not possible to do so even if we want to. In this case, however, $y=\sqrt{x}$ and we know that $\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$