How to Differentiate Solutions

Author: Indigo Curnick

Date: 2025-03-01

#mathematics   #science  


Differentiating Polynomials Solutions

  1. \(\frac{dy}{dx} = 4x\)
  2. \(\frac{dy}{dx} = 9x^2 + 8x -1\)
  3. \(\frac{dy}{dx} = -8x^3 -9x^2 - 7\)
  4. \(\frac{dy}{dx} = 25x^4 + 4x^3\)
  5. \(\frac{dy}{dx} - 0\)

Differentiation From First Principles Solutions

\[y = 3x\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{3(x + \Delta x) - 3x}{\Delta x}\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{3x + 3\Delta x - 3x}{\Delta x}\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{3 \Delta x}{\Delta x}\] \[\frac{dy}{dx} = 3\]

\[y = 2x^2 + 5x\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{2(x + \Delta x)^2 + 5(x + \Delta x) - 2x^2 - 5x}{\Delta x}\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{2(x^2 + 2x \Delta x + (\Delta x)^2) + 5x + 5\Delta x - 2x^2 - 5x}{\Delta x}\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{2x^2 + 4x \Delta x + 2(\Delta x)^2 + 5x + 5\Delta x - 2x^2 - 5x}{\Delta x}\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{4x \Delta x + 2(\Delta x)^2 + 5\Delta x}{\Delta x}\]

Notice how we can divide through by \(\Delta x\)

\[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} 4x + 2\Delta x + 5\] \[\frac{dy}{dx} = 4x + 5\]

\[y = x^3 - x\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{(x + \Delta x)^3 - x - \Delta x - x^3 + x}{\Delta x}\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{x^3 + 2x^2 \Delta x + x(\Delta x)^2 + x^2 \Delta x + 2x(\Delta x)^2 + (\Delta x)^3 - \Delta x - x^3}{\Delta x}\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{2x^2 \Delta x + x(\Delta x)^2 + x^2 \Delta x + 2x(\Delta x)^2 + (\Delta x)^3 - \Delta x}{\Delta x}\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} (2x^2 + x\Delta x + x^2 + 2x\Delta x + (\Delta x)^2 - 1)\]

\[\frac{dy}{dx} = 3x^2 - 1\]

\[y = c\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{c - c}{\Delta}\] \[\frac{dy}{dx} = 0\]

Finding Stationary Points with Differentiation Solutions

  1. Minimum at \(x= -1/4, y = -9/8\)
  2. Inflection at \(x=0, y = -2\)
  3. Minimum at \(x \approx -3.22366, y \approx -19.0593\), maximum at \(x \approx -0.961699, y \approx 12.0192\), minimum at \(x \approx 1.93536, y \approx -48.1278\)

Differentiation Product Rule Solutions

  1. \(\frac{dy}{dx} = x^2 (3 \cos(x) - x \sin(x))\)
  2. \(\frac{dy}{dx} = (4x - 1) \sin(x) + x (2x - 1) \cos(x)\)
  3. \(\frac{dy}{dx} = 5 e^x x^2 (x+3)\)
  4. \(\frac{dy}{dx} = 6x^2 (3 \ln(x) +1)\)
  5. \(\frac{dy}{dx} = e^x (\tan(x) + \sec^2 (x))\)

\[y = f(x) g(x)\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) g(x + \Delta x) - f(x) g(x)}{\Delta x}\]

The clever trick here is to add 0 in the form of \(-f(x)g(x + \Delta x) + f(x) g(x + \Delta x)\)

\[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) g(x + \Delta x) -f(x)g(x + \Delta x) + f(x)g(x + \Delta x) - f(x) g(x)}{\Delta x}\] \[\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \left( \frac{f(x + \Delta x) - f(x)}{\Delta x} \cdot g(x + \Delta x) + f(x) \cdot \frac{g(x + \Delta x) - g(x)}{\Delta x} \right)\]

Notice the following things

\[\lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = f^\prime(x)\] \[\lim_{\Delta x \rightarrow 0} g(x + \Delta x) = g(x)\] \[\lim_{\Delta x \rightarrow 0} f(x) = f(x)\] \[\lim_{\Delta x \rightarrow 0} \frac{g(x + \Delta x) - g(x)}{\Delta x} = g^\prime(x)\]

\[\frac{dy}{dx} = f^\prime (x) g(x) + g^\prime (x) f(x)\]

Differentiation Quotient Rule Solutions

  1. \(\frac{dy}{dx} = \frac{x \cos(x) - \sin(x)}{x^2}\)
  2. \(\frac{dy}{dx} = \frac{\sec^2(x) - \tan(x)}{e^x}\)
  3. \(\frac{dy}{dx} = \frac{-2x^2 + 4x + 1}{x^2 (2x + 1)^2}\)
  4. \(\frac{dy}{dx} = \frac{e^x (5x^2 -14x -2)}{x^3 (5x + 1)^2}\)

Define

\[h(x) = \frac{f(x)}{g(x)}\]

\[\frac{d}{dx} h(x) = \lim_{\Delta x \rightarrow 0} \frac{\frac{f(x + \Delta x)}{g(x + \Delta x)} - \frac{f(x)}{g(x)}}{\Delta x}\]

\[\frac{d}{dx} h(x) = \lim_{\Delta x \rightarrow 0} \frac{\frac{f(x + \Delta x) g(x) - f(x)g(x + \Delta x)}{g(x + \Delta x)g(x)}}{\Delta x}\] \[\frac{d}{dx} h(x) = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) g(x) - f(x)g(x + \Delta x)}{g(x + \Delta x)g(x) \Delta x}\]

\[\frac{d}{dx} h(x) = \lim_{\Delta x \rightarrow 0} \left( \frac{g(x)(f(x + \Delta x) - f(x))}{\Delta x g(x + \Delta x)g(x)} - \frac{f(x)(g(x + \Delta x) - g(x))}{\Delta x g(x + \Delta x)g(x)} \right)\] \[\frac{d}{dx} h(x) = \frac{g(x)f^\prime(x)}{(g(x))^2} - \frac{f(x)g^\prime(x)}{(g(x))^2}\] \[\frac{d}{dx} h(x) = \lim_{\Delta x \rightarrow 0} \frac{f^\prime(x)g(x) - f(x)g^\prime(x)}{(g(x))^2}\]

Differentiation Chain Rule Solutions

  1. \(\frac{dy}{dx} = 10 \sec^2 (10x + 4)\)
  2. \(\frac{dy}{dx} = 2 e^{x^2} x\)
  3. \(\frac{dy}{dx} = \sin(x) e^{1 - \cos(x)}\)
  4. \(\frac{dy}{dx} = \sin(x)(-\cos(2 + \cos(x)))\)

Assorted Differentiation Problems Solutions

Stationary Point Problems Solutions

  1. minimum at \(x = \frac{1}{\sqrt{2}}, y = \frac{1}{2} (1 + \ln(2))\)

  2. minimum at \(x \approx -1.83941, y \approx -34.8514\)

  3. maximum at \(x = -8, y = 450\), minimum at \(x=2, y=-50\)

Gradient Problems Solutions

  1. approx 13.54499...

  2. approx -19.0279...

  3. 195

  4. 94743

  5. \((2+e)\cos(1+e) \approx -3.9552...\)

Position of an Object Solutions

  1. \(\dot{x} = 4t(3t^2 - 9t + 20)\)
  2. \(\ddot{x} = 36t^2 - 72t + 80\)
  3. No, it just starts from rest at \(t=0\)

Population of Bacteria Solutions

  1. \(e^{10} \approx 22026.4657...\)
  2. \(e^{15} + 12 \approx 3269029.37247...\)

Competing Bacteria Solution

  1. \(b\) has the first peak
  2. \(b\) dies out first, \(a\) dies out last
  3. \(a\) starts with 2, \(b\) starts with 3