13. Chain Rule & Implicit Differentiation

The derivative of a composition is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. \[ (f\circ g)'(x) = f'(g(x)) g'(x) \]

Some of the problems in this tutorial, require the derivative of \(\ln\), the natural logarithm. Either ignore those problems, or use the fact that the derivative of \(\ln\) is: \[ \dfrac{d}{dx}\ln x=\dfrac{1}{x} \] In addition, some of the problems require the derivative of \(x^n\) where \(n\) is not a positive integer. Either ignore those problems, or use the fact that the derivative of \(x^n\) is: \[ \dfrac{d}{dx}x^n=n x^{n-1} \qquad \text{for all x} \]

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c. Tutorial on Chain Rule

Use the Chain Rule to differentiate the composite function:
\((f\circ g)(x)=\)

1. Identify the outer function: (Write it as a function of \(u\).)
   \(f(u)=\)   \(=\) \( \)

   \(f(u)\) is the outer function.

2. Identify the inner function:
   \(g(x)=\)   \(=\) \( \)

   \(g(x)\) is the inner function.

3. Find the derivative of the outer function: (Write it as a function of \(u\))
   \(f'(u)=\)   \(=\) \( \)

   \(f'(u)\) is the derivative of the outer function \(f(u)\) identified in question 1.

4. Find the derivative of the inner function:
   \(g'(x)=\)   \(=\) \(\)

   \(g'(x)\) is the derivative of the inner function \(g(x)\).

5. Find the derivative of the outer function evaluated at the inner function:
   \(f'(g(x))=\)   \(=\) \(\)

   \(f'(g(x))\) is \(f'(u)\) evaluated at \(u = g(x)\).

6. Find the derivative of the composite function.
   \((f\circ g)'(x)=\)   \(=\) \(\)

   \((f\circ g)'(x) = f'(g(x)) g'(x)\), the product \(f'(g(x))\) and \(g'(x)\).
   In words: "The derivative of a composition is the derivative of the outer function evaluated at the inner function times the derivative of the inner function."

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