Implicit Differentiation

This is how I understand implicit differentiation.

Say you want to take a derivative of an implicit function like \(x^2 + y^2 = 3\).

1. Take the derivative of everything with respect to \(x\):
   \(\frac{d}{dx} x^2 + \frac{d}{dx} y^2 = \frac{d}{dx} 3\)
   With a little simplification this is just \(2x + \frac{d}{dx} y^2 = 0\).
   
2. Cleverly apply the chain rule to get \(\frac{d}{dx} y^2\).
   Chain rule states that \(\frac{d}{du}\frac{du}{dx} = \frac{d}{dx}\).
   Define \(u = y^2\).
   By chain rule \(\frac{du}{dy}\frac{dy}{dx}\) = \(\frac{du}{dx}\).
   
3. Our formula is now \(2x + (2y)\frac{dy}{dx} = 0\). Time for some algebra!
   \(\frac{dy}{dx} = \frac{-2x}{2y} = \frac{-x}{y}\)
   

Technique based on the Chain Rule that allows diffrentiation of more functions.

EXAMPLE \(\frac{d}{dx} x^a = ax^{a-1}\)
This holds true for \(a=0, \pm 1, \pm 2\)…
What about fractional powers? Take \(a = \frac{m}{n}\)

\(y = x^{\frac{m}{n}}\) or \(y^n = x^m\)
We can apply derivative to equation 2 because the methods of diffrentiating the fractional exponent is unknown to us.

\(\frac{d}{dx} y^n = \frac{d}{dx} x^m\)

\(\big(\frac{d}{dy} y^n\big) \frac{dy}{dx}= \frac{d}{dx} x^m\)
or…
\(ny^{n-1} \frac{dy}{dx}= mx^{m-1}\)

$\frac{dy}{dx} =

\(x^2 + y^2 = 1\) is an implicit function, explicitly it is \(y = \pm \sqrt{1-x}\) (for convienience limit to positives for now).

Solving it explicitly:
\(y' = \frac{1}{2}()^{-1/2} (-2x)\)
NOTE: \(\frac{1}{2}()^{-1/2}\) = \(\frac{d}{d()}()^{-1/2}\)

Or implicitly:

• Diffrentiate function in the form \(x^2 + y^2 = 1\).
  
• \(\frac{d}{dx} (x^2 + y^2 = 1)\) or \(2x + 2yy' = 0\)
  
• Solve for y' which is \(\frac{-2x}{2y} = \frac{-x}{y}\) (solve algebraically).
  

Compare \(\frac{-x}{y}\) to explicit solution \(\frac{1}{2}(1-x^2)^{-1/2} (-2x) = \frac{-x}{\sqrt{1-x^2}}\) and find they are the same as \(y = \sqrt{1-x^2}\).

\(y^4 + xy^2 -2 = 0\)

• One can solve it explicitly by using the quadratic equation.
  
• Implicitly one can apply the product rule and the previous examples to diffrentiate this function.
  
• Writeup is left as an exercise for the reader.
  

\(f(x) = \sqrt{x}, g(y) = x, g(y) = y^2\)
NOTE: If \(y = f(x)\) and \(g(y)=x\), \(g(f(x)) = x\)

STATEMENT Implicit differentiation allows computing derivatives of any inverse function provided we know the derivative of the function.

EXAMPLE \(y = \tan^{-1}{x}\) and we'll use the equation \(\tan{y} = x\)

• Note that inverse functions are the function reflected over the line \(x=y\).
  
• Recall that derivative of tangent is \(\frac{d}{dy} \frac{\sin{y}}{\cos{y}} = \frac{1}{\cos^2{y}} = \sec^2{y}\).
  
• \(\frac{d}{dy} \tan{y} = 1\) or \(\sec^2{y} * y' = 1\).
  
• \(y' = \cos^2{y}\) which leads to \(\frac{d}{dx} \tan^{-1}{x} = \cos^2(\tan^{-1}{x})\).
  
  • Too complicated!
    
• Modelling as a right triangle and simplifying more yields \(\frac{d}{dx} \tan^{-1}{x} = \frac{1}{1+x^2}\).