Implicit differentiation enables you to find the derivative of y with respect to x without having to solve the original equation for y.
Implicit functions
Sometimes, equations involve variables that are intertwined and not easily separated. For example, \(y^{5}+3xy+x^{2}-5=0\) or \(y=\sin(xy))\).
In such expressions, \(y\) is said to be an implicit function of \(x\) as we cannot rearrange the expression to the form \(y=f(x)\). We can use implicit differentiation techniques to find \(\dfrac{dy}{dx}\) without having to solve the given equation for \(y\).
To do this, we need to use implicit differentiation. This process uses the chain rule because it is assumed that \(y\) is an unknown function of \(x\).
Using implicit differentiation
Consider a function \(f(y)\) which has intertwined variables \(x\) and \(y\). To differentiate the function with respect to \(x\), we use the chain rule:
In other words, to differentiate \(f(y)\) with respect to \(x\), we take the derivative with respect to \(y\), then multiply by the derivative with respect to \(x\) \(\left(\dfrac{dy}{dx}\right)\).
Example 1 – using implicit differentiation
Find \(\dfrac{d}{dx}(y^{2}x)\).
For this function, we can use the product rule first.
\[\begin{align*} \frac{d}{dx}(y^{2}x) & = \left(y^{2}\cdot\frac{d}{dx}(x)\right)+\left(x\cdot\frac{d}{dx}(y^{2})\right)\\
& = (y^{2}\cdot1)+\left(x\cdot\frac{d}{dx}(y^{2})\right)
\end{align*}\]
To find \(\dfrac{d}{dx}(y^{2})\), we need to use the chain rule. For every expression we differentiate that contains an \(x\) variable, we have to multiply by \(\dfrac{dy}{dx}\).
\[\begin{align*} \frac{d}{dx}(y^{2}) & = \frac{d}{dy}(y^{2})\times\frac{dy}{dx}(y^{2})\\
& = 2y\times\frac{dy}{dx}(y^{2})
\end{align*}\]
We can start by differentiating both sides of the equation with respect to \(x\), then rearrange the equation to find \(\dfrac{dy}{dx}\).
\[\begin{align*} \frac{d}{dx}(y^{3}) & = \frac{d}{dx}(2xy-7)\\
3y^{2}\cdot\frac{dy}{dx} & = 2(x\cdot\frac{dy}{dx}+y)\\
& = 2x\cdot\frac{dy}{dx}+2y\\
3y^{2}\cdot\frac{dy}{dx}-2x\cdot\frac{dy}{dx} & = 2y\\
(3y^{2}-2x)\frac{dy}{dx} & = 2y\\
\frac{dy}{dx} & = \frac{2y}{3y^{2}-2x}
\end{align*}\]
Find the value of the derivative at the point \((\pi,0)\) if \(\sin(xy)=2x\).
Differentiate each side with respect to \(x\). For \(\sin(xy)\), we use the product rule. Let \(u=xy\) and \(y=\sin(u)\). The derivative of \(y\) is simple.
\[y'=\cos(u)\]
To find \(u'\) is a bit more complex and involves the product rule.
\[\begin{align*} u' & = \frac{d}{dx}(xy)\\
& = y\cdot1+x\cdot\frac{dy}{dx}\\
& = y+x\frac{dy}{dx}
\end{align*}\]
The derivative of \(2x\) is simply \(2\).
Now, we can apply the chain rule.
\[\frac{d}{dx}(\sin(xy)) = \cos(xy)\times\left(y+x\cdot\frac{dy}{dx}\right)\]
To find the value of the derivative, we substitute in \(x=\pi\) and \(y=0\).
\[\begin{align*} \cos(xy)\times\left(y+x\cdot\frac{dy}{dx}\right) & = 2\\
\cos(\pi\times0)\times\left(0+\pi\cdot\frac{dy}{dx}\right) & = 2\\
1\times\left(\pi\cdot\frac{dy}{dx}\right) & = 2\\
\frac{dy}{dx} & = \frac{2}{\pi}
\end{align*}\]
Find the equation of the tangent line to the circle \(x^{2}+y^{2}=9\) at the point \((2,\sqrt{5})\).
At the point \((2,\sqrt{5})\), \(y'\) or the gradient \(m\) of the tangent is: \(m=-\dfrac{2}{\sqrt{5}}\).
To find the equation of the tangent, we can use \(y-y_{1}=m(x-x_{1})\). You might recall this from Linear functions.
\[\begin{align*} y-\sqrt{5} & = -\frac{2}{\sqrt{5}}(x-2)\\
y & = -\frac{2}{\sqrt{5}}x+\frac{4}{\sqrt{5}}+\sqrt{5}\\
& = -\frac{2}{\sqrt{5}}x-\frac{1}{\sqrt{5}}(4+5)\\
& = -\frac{2}{\sqrt{5}}x+\frac{9}{\sqrt{5}}
\end{align*}\]
Exercise – using implicit differentiation
Find \(\dfrac{d}{dx}\left(\dfrac{x}{y}\right)\). Hint: use quotient rule.
