The chain rule

\(\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\)

Examples

Find the derivative of the following:
a) \(y=u^5\)\(u=1-x^3\)
\(y=(1-x^3)^5 \Longrightarrow \frac{\partial y}{\partial x}=5(1-x^3)^4\cdot -3x^2\Longrightarrow \frac{\partial y}{\partial x}=-15x^2(1-x^3)^4\)
b) \(y=\frac{10}{(x^2+4x+5)^7}\)
\(\frac{\partial y}{\partial x}=-70(x^2+4x+5)^{-8}\cdot (2x+4)\Longrightarrow \frac{\partial y}{\partial x}=-70(2x+4)(x^2+4x+5)^{-8}\)
c) \(y=\left(\frac{x-1}{x+3}\right)^{1/3}\)
\(\frac{\partial y}{\partial x}=\frac{1}{3}(\frac{x-1}{x+3})^{-2/3}\cdot \frac{(x+3)-(x-1)}{(x+3)^2}\)

Alternative formulation of the chain rule

Proof of the chain rule

  1. Write \(F(x)=f(g(x))\)
  2. Use Newton's quotient to write \(F^{'}(x)=\lim_{h\to 0} \frac{f(g(x+h))-f(g(x))}{h}\)
  3. Define \(k=g(x+h)-g(x)\) and substitute \(g(x+h)=g(x)+k\) into \(f(g(x+h))\)
  4. Multiply and divide \(F^{'}(\cdot)\) by \(k\)\(F^{'}(x)=\lim_{h\to 0} \frac{f(g(x)+k)-f(g(x))}{k}\cdot \frac{k}{h}=F^{'}(x)=\lim_{h\to 0} \frac{f(g(x+h))-f(g(x))}{k}\cdot \frac{g(x+h)-g(x)}{h}\)
  5. Taking the limit yields: \(F^{'}(\cdot)=f^{'}(g(x))\cdot g^{'}(x)\)

Higher order derivatives

Examples

Find the second derivative of the following equations
a) \(Y=AK^{\alpha}\)
\(Y^{'}=\alpha AK^{\alpha-1}\)
\(Y^{''}=\alpha(\alpha-1)AK^{\alpha-2}\)
b) \(L=\frac{t}{t+1}, t\ge 0\)
\(L^{'}=\frac{1}{(t+1)^{2}}\)
\(L^{''}=-2(t+1)^{-3}\)

Convex and concave functions