Chap 23 Review

\[ \newcommand{\dnorm}{\text{dnorm}} \newcommand{\pnorm}{\text{pnorm}} \newcommand{\recip}{\text{recip}} \]

Exercise 1 What’s the derivative of \(x^3\) with respect to \(x\)? Solve this by writing \(x^3\) as \(x \cdot x^2\) and applying the product rule.

Since we already know \(\partial_x x\) (it is 1) and \(\partial_x x^2\) (it is \(2x\)) let’s apply the product rule to find \(\partial_x x^3\): \[\partial_x [\color{magenta}x \times \color{brown}{x^2}] = \color{magenta}{[\partial_x x]} \times \color{brown}{x^2} \ + \color{magenta}x \times \color{brown}{[\partial_x x^2]} =\color{magenta}1\times \color{brown}{x^2} + \color{magenta}x \times \color{brown}{2x} = 3 x^2\]

Exercise 4 Recognizing \(e^{2x}\) as \(e^x \times e^x\), apply the product rule.

The functions being multiplied are identical: \(f(x) \equiv e^x\) and \(g(x) \equiv e^x\).

Naturally their derivatives are also identical, and since \(e^x\) is involved they will both be \(e^x\):

\[f'(x) = e^x\ \ \ \ \text{and} \ \ \ \ g'(x) = e^x \ .\]

The formula for the product rule involves \(f'() g()\) and \(g'() f()\). Each of these are \(e^x\ e^x\) which is the same as \(e^{2x}\). The sum \(f'() g() + g'() f()\) is therefore \(2 e^{2x}\).

Exercise 2 Use the chain rule to find the derivative \(\partial_x e^{2x}\).

Hint: The first step is to identify the interior \(f()\) and the exterior \(g()\) functions. Then differentiate each to get \(f'()\) and \(g()\) and apply the formula.

\(g(x) \equiv 2x\) is the interior function in \(e^{2x}\) and \(f(x) \equiv \exp(x)\) is the exterior function. Thus \[\partial_x e^{2x} = f'\left(\strut g(x)\right) g'(x) = \exp\left(\strut g(x)\right) 2 = 2 e^{2x}\ .\]

Exercise 3  

In draft

Make this an exercise in three parts: do the derivative using power-law rules, then using multiplication, then using composition.

There is one family of functions for which function composition accomplishes same thing as multiplying functions: the power-law family.

Consider, for instance, the function \(h(x) \equiv \left[3x\right]^4\). Let’s let \(g(x) \equiv 3x\) and \(f(y) \equiv y^4\). With these definitions, \(h(x) = f(g(x))\).

Recognizing that \(\partial_y f(y) = 4 y^3\) and \(\partial_x g(x) = 3\), the chain rule gives \[\partial_x h(x) = \underbrace{4 g(x)^3}_{f'(g(x))} \times \underbrace{3}_{g'(x)} = \underbrace{4 (3 x)^3}_{f'(g(x))} \times 3 = 4\cdot 3^4 \times x^3 = 324\ x^3\] Another way to look at the same function is \(g(x)\) multiplied by itself 3 times: \[h(x) = g(x)\cdot g(x) \cdot g(x) \cdot g(x)\] This is a product of 4 terms. Applying the product rule gives \[\begin{eqnarray} \partial_x h(x) &=& \color{blue}{g'(x)}\cdot g(x)\cdot g(x) \cdot g(x) +\\ &\ & g(x)\cdot \color{blue}{g(x)}'\cdot g(x) \cdot g(x) +\\ &\ & g(x)\cdot g(x)\cdot \color{blue}{g(x)'} \cdot g(x) +\\ &\ & g(x)\cdot g(x)\cdot g(x) \cdot \color{blue}{g'(x)} \end{eqnarray}\] Since multiplication is commutative, all four terms are the same, each being \(3^4 x^3\). The sum of all four is therefore \(4 \times 3^4 x^3 = 324 x^3\).

These are two long-winded ways of getting to the result. For most people, differentiating power-law functions algebraically is simplified by using the rules of exponentiation rather than the product or chain rule. Here, \[h(x) \equiv \left[3x\right]^4 = 3^4 x^4\]so \(\partial_x h(x)\) is easily handled as a scalar (\(3^4\)) times a function \(x^4\). Consequently, applying the rule for differentiating power laws, \[\partial_x h(x) = 3^4 \times \partial_x x^4 = 3^4 \times 4 x^3 = 324 x^3\] As another example, take \(h(x) \equiv \sqrt[4]{\strut x^3}\). This is, of course, the composition \(f(g(x))\) where \(f(y) \equiv y^{1/4}\) and \(g(x) \equiv x^3\). Applying the chain rule to find \(\partial_x h(x)\) will work (of course!), but is more work than applying the rules of exponentiation followed by a simple power-law differentiation. \[h(x) = \sqrt[4]{\strut x^3} = x^{3/4}\ \ \text{so}\ \ \partial_x h(x) = \frac{3}{4} x^{(3/4 - 1)} = \frac{3}{4} x^{-1/4}\]

No answers yet collected