Short and Sweet Calculus

## 3.8 Implicit Differentiation

Sometimes we have to deal with equations of the form $F(x,y)=0,$ where the equation is not solved for $$y$$ such as

$x^{2}+y^{2}=4,$ and $x^{3}-3xy+y^{3}=0.$ In such cases, we say $$y$$ is implicitly expressed in terms of $$x$$; if a value of $$x$$ is given, a value or values of $$y$$ may be determined. Sometimes, we can solve these equations for $$y$$ in terms of $$x$$, thereby $$y$$ becomes an explicit function (or perhaps several functions) of $$x$$.

Sometimes it is very difficult or even impossible to solve an implicit relation for $$y$$. In such cases, if we assume that $$y$$ can be defined as one or more differentiable functions of $$x$$, we can apply the chain rule to find $$dy/dx$$ directly without solving the equation. In this method that is known as implicit differentiation, we differentiate both sides of an equation with respect to $$x$$ and treat $$y$$ as a differentiable function of $$x$$. Then, we try to solve for $$dy/dx$$.

Example 3.17. If $$x^{2}+y^{2}=4$$, find $$dy/dx$$ using implicit differentiation.

Solution

We assume $$dy/dx$$ exists and apply $$\dfrac{d}{dx}$$ to both sides of the equation: \begin{aligned} \frac{d}{dx}(x^{2}+y^{2}) & =\frac{d}{dx}4\\ \frac{d}{dx}x^{2}+\frac{d}{dx}y^{2} & =0\tag{\small Derivative of a constant is zero}\\ 2x+2y\frac{dy}{dx} & =0.\tag{i}\end{aligned} For the last step, we have used the chain rule: $\frac{d}{dx}(y^{2})=\frac{d}{dy}(y^{2})\cdot\frac{dy}{dx}=2y\frac{dy}{dx}.$ We can solve Equation (i) for $$dy/dx$$ and obtain: $\frac{dy}{dx}=-\frac{x}{y}.$

• Note that in implicit differentiation only those values of the variables which satisfy the original relation can be substituted in the derivative.

Example 3.18. Find the equation of the tangent line to the curve described by $$x^{3}+y^{3}-3xy=0$$ at $$(\sqrt{2},\sqrt{4})$$.

Solution

First of all, we note that $$x=\sqrt{2}$$ and $$y=\sqrt{4}$$ satisfy $$x^{3}+y^{3}-3xy=0$$. Differentiating with respect to $$x$$, we obtain: \begin{aligned} \frac{d}{dx}(x^{3})+\frac{d}{dx}\left(y^{3}\right)-3\frac{d}{dx}(xy) & =0,\\ 3x^{2}+3y^{2}\frac{dy}{dx}-3y-3x\frac{dy}{dx} & =0,\\ \frac{dy}{dx}\left(3y^{2}-3x\right) & =3(y-x^{2})\\ \frac{dy}{dx} & =\frac{y-x^{2}}{y^{2}-x}.\end{aligned} The slope of the tangent at $$(\sqrt{2},\sqrt{4})$$ is thus $y’=\left.\frac{y-x^{2}}{y^{2}-x^{2}}\right|_{x=\sqrt{2},y=\sqrt{4}}=\frac{\sqrt{4}-(\sqrt{2})^{2}}{\sqrt{16}+\sqrt{4}}=\frac{\sqrt{4}-\sqrt{4}}{\sqrt{16}+\sqrt{4}}=0.$ This shows the tangent to this curve is horizontal and its equation becomes $$y=\sqrt{4}$$. The following figure shows the curve described by the equation $$x^{3}+y^{3}-3xy=0$$ and its tangent at the point $$(\sqrt{2},\sqrt{4})\approx(1.26,1.587)$$. The curve described by $$x^{3}+y^{3}-3xy=0$$ and its tangent at $$(\sqrt{2},\sqrt{4})$$.