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DA4. Differentiation Rules

Statement

Write an example of a function whose derivative can be found by using the following rules:

  1. Product rule and special function differentiation rules.
  2. Power rule, quotient rule, and chain rule.
  3. Chain rule twice.
  4. Implicit differentiation and special function differentiation rule.

Solution

1. Product rule and special function differentiation rules

According to (khan Academy, n.d.), the product rule states that derivative of a product of two functions the sum of the each function multiplied by the derivative of the other function. In other words, if \(f(x) = g(x)h(x)\), then \(f'(x) = g'(x)h(x) + g(x)h'(x)\).

The special function differentiation rules dictates how to find the derivative of special functions such as trigonometric, exponential, and logarithmic functions. For example, the derivative of \(\sin(x)\) is \(\cos(x)\), the derivative of \(\cos(x)\) is \(-\sin(x)\) and the derivative of \(\ln(x)\) is \(\frac{1}{x}\) (HMC, n.d.).

As a function that matches both rules, we can use \(f(x) = x^2 ln(x)\) as an example.

\[ \begin{aligned} f(x) &= x^2 ln(x) \\ f'(x) &= (x^2)' ln(x) + x^2 (ln(x))' \\ &= 2x ln(x) + x^2 \frac{1}{x} \\ &= 2x ln(x) + x \end{aligned} \]

2. Power rule, quotient rule, and chain rule

According to (MathIsFun, n.d.),

  • The power rule states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).
  • The quotient rule states that if \(f(x) = \frac{g(x)}{h(x)}\), then \(f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}\).
  • The chain rule states that if \(f(x) = g(h(x))\), then \(f'(x) = g'(h(x))h'(x)\).

As an example that matches the three criteria, I’ll choose: \(g(x) = x^2\), \(h(x) = \ln(x)\), \(k(x)=x+1\) and \(f(x) = g(\frac{k(x)}{h(x)})\).

\[ \begin{aligned} f(x) &= g(\frac{k(x)}{h(x)}) \\ &= g(\frac{x+1}{\ln(x)}) \\ &= (\frac{x+1}{\ln(x)})^2 \\ f'(x) &= g'(\frac{k(x)}{h(x)}) (\frac{k(x)}{h(x)})' \\ &= g'(\frac{x+1}{\ln(x)}) (\frac{x+1}{\ln(x)})' \\ &= 2(\frac{x+1}{\ln(x)}) (\frac{k'(x)h(x) - h'(x) k(x)}{h(x)^2}) \\ &= 2(\frac{x+1}{\ln(x)}) (\frac{(1)\ln(x) - \frac{1}{x}(x+1)}{\ln^2(x)}) \\ &= 2(\frac{x+1}{\ln(x)}) (\frac{\ln(x) - \frac{x+1}{x}}{\ln^2(x)}) \\ &= 2(\frac{x+1}{\ln(x)}) (\frac{x\ln(x) - (x + 1)}{x\ln^2(x)}) \\ &=\frac{2(x+1)(xln(x) - (x+1))}{xln^3(x)} \end{aligned} \]

It can be simplified further, but I’ll leave it as is.

3. Chain rule twice

Okay, suppose that we have \(h(x)=x^2\) and \(g(x)=\ln(x)\) and \(k(x)=x+1\) and \(f(x)=h(g(k(x)))\). Then:

\[ \begin{aligned} f(x) &= h(g(k(x))) \\ f'(x') &= h'(g(k(x))) . (g(k(x)))' \\ &= h'(g(k(x))) . g'(k(x)) . k'(x) \\ &= 2g(k(x)) . \frac{1}{k(x)} . 1 \\ &= \frac{2g(k(x))}{k(x)} \\ &= \frac{2\ln(x+1)}{x+1} \end{aligned} \]

4. Implicit differentiation and special function differentiation rule

I’ll take \(x^2 + ln(y) = 1\) as an example.

\[ \begin{aligned} x^2 + ln(y) &= 1 \\ \frac{d}{dx} (x^2 + ln(y)) &= \frac{d}{dx} (1) \\ \frac{d}{dx} (x^2) + \frac{d}{dx} (ln(y)) &= 0 \\ 2x + \frac{1}{y} \frac{dy}{dx} &= 0 \\ \frac{dy}{dx} &= -2xy \end{aligned} \]

References