Example 1 Consider :y + x + 5 = 0 \,. This equation is easy to solve for , giving :y = -x - 5 \,, where the right side is the explicit form of the function . Differentiation then gives . Alternatively, one can totally differentiate the original equation: :\begin{align} \frac{dy}{dx} + \frac{dx}{dx} + \frac{d}{dx}(5) &= 0 \, ; \\[6px] \frac{dy}{dx} + 1 + 0 &= 0 \,. \end{align} Solving for gives :\frac{dy}{dx} = -1 \,, the same answer as obtained previously.
Example 2 An example of an implicit function for which implicit differentiation is easier than using explicit differentiation is the function defined by the equation : x^4 + 2y^2 = 8 \,. To differentiate this explicitly with respect to , one has first to get :y(x) = \pm\sqrt{\frac{8 - x^4}{2}} \,, and then differentiate this function. This creates two derivatives: one for and another for . It is substantially easier to implicitly differentiate the original equation: :4x^3 + 4y\frac{dy}{dx} = 0 \,, giving :\frac{dy}{dx} = \frac{-4x^3}{4y} = -\frac{x^3}{y} \,.
Example 3 Often, it is difficult or impossible to solve explicitly for , and implicit differentiation is the only feasible method of differentiation. An example is the equation :y^5-y=x \,. It is impossible to
algebraically express explicitly as a function of , and therefore one cannot find by explicit differentiation. Using the implicit method, can be obtained by differentiating the equation to obtain :5y^4\frac{dy}{dx} - \frac{dy}{dx} = \frac{dx}{dx} \,, where . Factoring out shows that :\left(5y^4 - 1\right)\frac{dy}{dx} = 1 \,, which yields the result :\frac{dy}{dx}=\frac{1}{5y^4-1} \,, which is defined for :y \ne \pm\frac{1}{\sqrt[4]{5}} \quad \text{and} \quad y \ne \pm \frac{i}{\sqrt[4]{5}} \,. ==References==