implicit differentiation
Implicit differentiation
Sometimes the relationship between \[y\] and \[x\] is not explicit; rather, it is implicit. For instance, we might know that \[x^{2}-y=4\]. This equality defines a relationship between \[x\] and \[y\]; if we know \[x\], we could figure out \[y\]. However, the implicit relationship between \[x\] and \[y\] is can be complicated and example would be \[\sin y+y^{3}+x^{3}-6=0\]. In this case there is no way to solve for \[y\] in terms of elementary functions, but it still can be solved with implicit differentiation.
The key concept to here is to find \[y^{\prime}\] even if we don't exactly know how \[y\] and \[x\] relate.
Concept
We will be utilising the chain rule in a rather smart manner. The definition of chain rule is given as \[f^{\prime}(g(x))\cdot g^{\prime}(x)\]. Suppose \[y\] relates to \[x\] with function \[g\], that would mean \[y=g(x)\], thus we can rewrite the chain rule as \[f^{\prime}(y)\cdot y^{\prime}\].
Using the example above, we can rewrite the equation as \[\sin y+y^{3}=-x^{3}+6\], now differentiating both sides of the equation with respect to \[x\], we'll get \[\left( \sin y+y^{3} \right)^{\prime}=\left( -x^{3}+6 \right)^{\prime}\]. Now, differentiating the right-side of the equation is relatively simple, \[\left( -x^{3}+6 \right)^{\prime}=-3x^{2}\]. The interesting part is in differentiating the left-side of the equation. Now, according to the linearity of differentiation, \[\left( \sin y+y^{3} \right)^{\prime}\] can be written as \[\left( \sin y \right)^{\prime}+\left( y^{3} \right)^{\prime}\].
We'll now go term by term. \[(\sin y)^{\prime}\] or, in Leibniz notation, \[\frac{d}{dx}(\sin y)\] (do note that \[\frac{d}{dx}\] is valid as \[y\] is defined as \[g(x)\], making \[\frac{d}{dx}(\sin y)\] as shorthand for \[\frac{d}{dx}(\sin g(x))\]), equals to \[(\sin y)^{\prime}=(\sin g(x))^{\prime}=\cos y\cdot y^{\prime}\] according to chain rule. Applying the same process to the term \[y^{3}\], we get \[\left( y^{3} \right)^{\prime}=\left( (y)^{3} \right)^{\prime}=3(y)^{2}\cdot y^{\prime}\]. Notice the brackets? The reason is that we treat this as \[f(g(x))\] where \[f(x)=x^{3}\] and \[g(x)=y\] to fulfill the requirements to apply chain rule.
Combining everything together, we get the derivative for the equation as \[\cos y\cdot y^{\prime}+3y^{2}\cdot y^{\prime}=-3x^{2}\]. Solving for \[y^{\prime}\] we get \[y^{\prime}=\frac{dy}{dx}=\frac{-3x^{2}}{\cos y+3y^{2}}\].