The Derivative of Implicit Functions

As you probably know the implicit function is defined as \(f(x,y) = 0. \) In implicit function the dependency between \(x\) and derivative function \(y\) is defined as \(f(x,y) = 0\) then the calculation of derivative \(y_x' = \frac{dy}{dx}\) is calculated as:

1. Calculate the derivative with respect to \(x\) by considering the \(y\) as the function of argument \(x\),
2. Equate that derivative with zero i.e. $$\frac{d}{dx} f(x,y) = 0, $$ and
3. Solve the equation \(\Rightarrow\) determine the \(y' = \frac{dy}{dx}.\)

• Example 1 Determine the \(y'\) of the implict function \(x^3 + y^3- 3axy = 0\)
  Solution \begin{eqnarray}x^3 + y^3- 3axy = 0/'\\ 3x^2 + 3y^2y' - 3ay - 3axy') &=& 0/:3 x^2 - ay + y'(y^2 - ax) &=& 0 \\ y'(y^2-ax) &=& ay - x^2 /:(y^2-ax) \\ y' &=& \frac{x^2 -ay }{ax - y^2}\end{eqnarray}
• Example 2 Determine the \(y'\) of the implict function \(2x - 5y + 10 = 0\)
  Solution \begin{eqnarray}2x - 5y + 10 &=& 0/' \\ 2 - 5y' &=& 0\\ 5y' &=& 2/:5 \\ y' &=& \frac{2}{5}\end{eqnarray}
• Example 3 Determine the \(y'\) of the implict function \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 0\)
  Solution \begin{eqnarray}\frac{x^2}{a^2} + \frac{y^2}{b^2} &=& 0/' \frac{2x}{a^2} + \frac{2yy'}{b^2} &=& 0\\ \frac{2yy'}{b^2} &=& - \frac{2x}{a^2} \\ y' &=& - \frac{b^2 x}{a^2 y}\end{eqnarray}
• Example 4 Determine the \(y'\) of the implicit function \(x^3 + y^3 = a^3,\) where a is a constant.
  Solution \begin{eqnarray}x^3 + y^3 &=& a^3/' \\ 3x^2 + 3y^2 y' &=& 0 \\ y^2 y' &=& -x^2/:y^2 \\ y' &=& -\frac{x^2}{y^2}\end{eqnarray}
• Example 5 Determine the \(y'\) of the implicit function \(x^3 + x^2 y + y^2 = 0\)
  Solution \begin{eqnarray}x^3 + x^2 y + y^2 &=& 0/' \\ 3x^2 + 2xy + x^2 y' + 2y y' &=& 0 \\ y'(x^2 + 2y) &=& - 3x^2 - 2xy/: (x^2 +2y) \\ y' &=& -\frac{3x^2 + 2xy}{x^2 + 2y}\end{eqnarray}
• Example 6 Determine the \(y'\) of the implicit function \(\sqrt{x} + \sqrt{y} = \sqrt{a},\) where a is a constant.
  Solution \begin{eqnarray}\sqrt{x} + \sqrt{y} &=& \sqrt{a}\\ \frac{1}{2}\frac{1}{\sqrt{x}} + \frac{1}{2}\frac{1}{\sqrt{y}} y' &=& 0/\cdot 2 \\ \frac{1}{\sqrt{y}}y' &=& -\frac{1}{\sqrt{x}}/\cdot \sqrt{x}\\ y' &=& -\sqrt{\frac{y}{x}}\end{eqnarray}
• Example 7 Determine the \(y'\) of the implicit funciton \(\sqrt[3]{x^2} + \sqrt[3]{y^2} = \sqrt[3]{a^2}\)
  Solution \begin{eqnarray}\sqrt[3]{x^2} + \sqrt[3]{y^2} &=& \sqrt[3]{a^2}/'\\ \frac{1}{\sqrt[3]{x}} + \frac{1}{\sqrt[3]{y}} y' &=& 0 \\ y' &=& -\sqrt[3]{\frac{y}{x}}\end{eqnarray}
• Example 8
  Solution \begin{eqnarray}\end{eqnarray}
• Example 9 Determine the \(y'\) of the implicit function \(y - 0.3\sin y = x\)
  Solution \begin{eqnarray}y - 0.3\sin y &=& x/'\\y' - 0.3\cos y y' &=& 1 \\ y' = \frac{1}{1-0.3\cos y}\end{eqnarray}
• Example 10 Determine the \(y'\) of the implicit function \(2\cos(x+y) - b= 0. \)
  Solution \begin{eqnarray}a\cos^2(x+y) &=& b/'\\ 2\cos(x+y) (1+y') &=& 0\\ y' &=& \frac{1-2\cos(x+y)}{2\cos(x+y)} \end{eqnarray}