Derivative Summary
This lesson is meant to summarize all of the main concepts covered in the Derivatives and Types of Derivatives units. As we have previously discussed, there are several different rules for determining the derivatives of an equation or function. The tables below will help summarize these rules:
Basic Derivatives
| Rule | Original | Derivative |
|---|---|---|
| Constant | \(f(x) = c\) | \(f'(x) = 0\) |
| Power | \(f(x) = x^n\) | \(f'(x) = nx^{n-1}\) |
| Constant Multiple | \(f(x) = K\cdot g(x)\) | \(f'(x) = K\cdot g'(x)\) |
| Sum | \(f(x) = g(x) + H(x)\) | \(f'(x) = g'(x) + H'(x)\) |
| Difference | \(f(x) = g(x) - H(x)\) | \(f'(x) = g'(x) - H'(x)\) |
| Product | \(f(x) = g(x)H(x)\) | \(f'(x) = g(x)\cdot H'(x) + H(x)\cdot g'(x)\) |
| Quotient | \(f(x) = \cfrac{g(x)}{H(x)}\) | \(f'(x) = \cfrac{H(x)\cdot g'(x) - g(x)\cdot H(x)}{[H(x)]^2}\) |
| Chain | \(f(x) = g(H(x))\) | \(f'(x) = g'(H(x))\cdot H'(x)\) |
Exponential and Logarithm Derivatives
| Exponential | \(f(x) = a^x\) | \(f'(x) = a^x\cdot \text{ln}a\) |
| Natural Exponential | \(f(x) = e^x\) | \(f'(x) = e^x\) |
| Logarithm | \(f(x) = \text{log}_b(x)\) | \(f'(x) = \cfrac{1}{x\cdot \text{ln}(b)}\) |
| Natural Logarithm | \(f(x) = \text{ln}(x)\) | \(f'(x) = \cfrac{1}{x}\) |
Trigonometric Derivatives
| Sine | \(f(x) = \sin(x)\) | \(f'(x) = \cos(x)\) |
| Cosine | \(f(x) = \cos(x)\) | \(f'(x) = -\sin(x)\) |
| Tangent | \(f(x) = \tan(x)\) | \(f'(x) = \cfrac{1}{\cos^2(x)} = \sec^2(x)\) |
| Cotangent | \(f(x) = \cot(x)\) | \(f(x) = -\cfrac{1}{\sin^2(x)} = -\csc^2(x)\) |
| Secant | \(f(x) = \sec(x)\) | \(f'(x) = \tan(x)\sec(x)\) |
| Cosecant | \(f(x) = \csc(x)\) | \(f'(x) = -\cot(x)\csc(x)\) |
\(y = \sin(x^2)\cos(5^x)\)
In order to differentiate this expression, we can use Product Rule.
First, we can identify \(f(x)\) and \(g(x)\):
\(\textcolor{red}{f(x) = \sin(x^2)}\)
\(\textcolor{blue}{g(x) = \cos(5^x)}\)
In order to determine \(f'(x)\), we can use Chain Rule:
\(\textcolor{red}{f'(x) = \cos(x^2)(2x)}\)
In order to determine \(g'(x)\), we can use Chain Rule:
\(\textcolor{blue}{g'(x) = -\sin(5^x)(5^x\cdot \text{ln}(5))}\)
We can now use Product Rule to differentiate this expression:
\(y' = \textcolor{red}{f'(x)}\textcolor{blue}{g(x)} + \textcolor{blue}{g'(x)}\textcolor{red}{f(x)}\)
\(y' = \textcolor{red}{\cos(x^2)(2x)}\textcolor{blue}{\cos(5^x)} + [\textcolor{blue}{-\sin(5^x)(5^x\cdot \text{ln}(5))}\textcolor{red}{\cos(x^2)(2x)}]\)
\(y' = \cos(x^2)(2x)\cos(5^x) - \sin(5^x)(5^x\cdot \text{ln}(5))\cos(x^2)(2x)\)
Therefore, we can determine that \(\boldsymbol{y' = \cos(x^2)(2x)\cos(5^x) -\sin(5^x)(5^x\cdot \text{ln}(5))\cos(x^2)(2x)}\).
\(y = (\sin(5x+e^x)^4)\)
First, we can determine \(f'(g(x))\). This involves using Chain Rule twice. We can use it to multiply the original function by its exponent (\(4\)):
\(= 4[\sin(5x+e^x)]^3\)
We can then use it to multiply the function by its trignometric derivative, \(\cos\):
\(\cos(5x+e^x)\)
When we put these 2 portions together, we get \(f'(g(x))\):
\(\textcolor{red}{f'(g(x)) = 4[\sin(5x+e^x)]^3\cos(5x+e^x)}\)
Next, we can determine \(g'(x)\). This involves using Power Rule and the Natural Exponential Formula:
\(\textcolor{blue}{g'(x) = 5+e^x}\)
Finally, we can use the Chain Rule to find the derivative of the whole expression:
\(y' = \textcolor{red}{f'(g(x))}\cdot \textcolor{blue}{g'(x)}\)
\(y' = \textcolor{red}{4[\sin(5x+e^x)]^3\cos(5x+e^x)}[\textcolor{blue}{5+e^x}]\)
Therefore, we can determine that \(\boldsymbol{y' = 4[\sin(5x+e^x)]^3\cos(5x+e^x)[5+e^x]}\).
First, we can substitute \(\cfrac{e}{2}\) for \(x\) in the original expression to determine its respective \(y\)-coordinate:
\(y \left(\cfrac{e}{2}\right) = \text{ln}(2) \left(\cfrac{e}{2}\right)\)
\(y \left(\cfrac{e}{2}\right) = \text{ln}(e)\)
\(y \left(\cfrac{e}{2}\right) = 1\)
Therefore, we can determine that the Point is \(\left(\cfrac{e}{2}, 1\right)\).
Next, we can differentiate the original expression in order to find the equation of its tangent:
\(y' = \left(\cfrac{1}{2x}\right)(2)\)
\(y' = \cfrac{1}{x}\)
Then, we can substitute \(\cfrac{e}{2}\) for \(x\) to determine the tangent's slope:
\(y\left(\cfrac{e}{2}\right) = \cfrac{1}{e/2}\)
\(m = \cfrac{1}{e/2}\cdot \cfrac{2}{2}\)
\(m = \cfrac{2}{e}\)
Finally, we can substitute all the pertinent values into the slope-intercept equation to determine the \(y\)-intercepts:
\(y = mx + b\)
\(1 = \left(\cfrac{2}{e}\right) \left(\cfrac{e}{2}\right) + b\)
\(1 = 1 + b\)
\(b = 0\)
\(y = \cfrac{2}{e}x\)
Therefore, we can determine that the equation of the tangent is \(\boldsymbol{y = \cfrac{2}{e}x}\).
Given this information, we can sketch a graph of the original expression and its tangent at \(x = \cfrac{e}{2}\):
If \(y = te^t - e^t - 2t^2\) represents the movement (distance to the origin) of a particle along a straight line:
- When does velocity equal to \(0\)?
- Is there a maximum or minimum distance to the origin?
- Determine the acceleration function.
- Is there a maxmimum or minimum velocity?
i. First, we can differentiate the original function in order to determine the velocity function:
\(y' = 1e^t + te^t - e^t - 4t\)
\(y' = te^t - 4t\)
Next, we can set the function equal to \(0\). We can then factor the function to determine its roots:
\(0 = te^t - 4t\)
\(0 = t(e^t - 4)\)
\(t = 0\)
In order to determine the root of \(e^t - 4\), we can set the 2 terms equal to each other. We can then find the \(ln\) of \(4\) to determine solve for that root:
\(e^t - 4 = 0\)
\(e^t = 4\)
\(t = \ln4\)
\(t = 1.39\)
Therefore, we can determine that the velocity equals \(0\) when \(t = 0\;[\text{s}]\) or \(t = 1.39\;[\text{s}]\).
ii. In order to determine if there is a maximum or minimum distance to the origin, we can use the First Derivatives Test using the zeroes that we determined in the last question:
| Interval | Test \(x\)-value | \(f'(x)\) | Conclusion |
|---|---|---|---|
| \(\textcolor{green}{(-∞, 0)}\) | \(-1\) | \(f'(-1)= 3.63\) | \(f\) is Increasing |
| \(\textcolor{red}{(0, 1.39)}\) | \(1\) | \(f'(1)= -1.28\) | \(f\) is Decreasing |
| \(\textcolor{green}{(1.39, ∞)}\) | \(2\) | \(f'(2)= 6.78\) | \(f\) is Increasing |
Therefore, we can determine that there is a minimum at \(\boldsymbol{t = 1.39\;[\textbf{s}]}\).
iii. In order to determine the acceleration, we can differentiate the velocity function:
\(\boldsymbol{f'' = 1e^t + te^t - 4}\)
iv. In order to determine if there is a max/min velocity, we can factor the acceleration function to help determine its zeroes:
\(f'' = e^t(1 + t - 4e^{-t})\)
As you can see, when we factored \(e^t\) out of the function, we multiplied \(4\) by \(e^{-t}\). This is because these values will end up cancelling each other out.
As \(e^t\) will never equal \(0\), we can determine if \(1 + t\) will intersect with the exponential (\(4e^{-t}\)) to determine if there is a max/min. We can determine this more easily by creating a table of values:
| x-value | -1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|
| 1 + t | 0 | 1 | 2 | 3 | 4 |
| 4e⁻ᵗ | 10.87 | 4 | 1.47 | 0.54 | 0.199 |
Based on the table of values, we can determine that the functions intersect somewhere between \(x = 0\) and \(x = 1\). We can sketch a graph to get a visual indication of where these functions intersect:
Therefore, we can determine that there is a max/min value around \(\boldsymbol{x = 0.8}\).
When a particular medication is swallowed by a patient, the concentration of the active ingredient, in parts per million, in the bloodstream is given by the equation \(C(t) = 150t(0.5)^t\) after \(t\) hours.
- What is the highest concentration of the medication?
- How fast is the concentration decreasing after \(2\) hours?
i. First, we can determine the derivative of the original equation:
\(C'(t) = 150(0.5)^t + 150t(0.5)^t\cdot \ln(0.5)\)
Next, we can factor this equation in order to determine its zeroes more easily:
\(C'(t) = 150(0.5)^t[1 + t\cdot \ln(0.5)]\)
We can determine that since \(150\) and \((0.5)^t\) can never equal \(0\) that they can't count as Critical Points. As a result, we can determine the Critical Point of \(1 + t\cdot \ln(0.5)\):
\(0 = 1 + t\cdot \ln(0.5)\)
\(t\cdot \ln(0.5) = -1\)
\(\cfrac{t\cdot \cancel{\ln(0.5)}}{\cancel{\ln(0.5)}} = \cfrac{-1}{\ln(0.5)}\)
\(t = 1.44\)
We can now use the First Derivatives Test to determine the highest concentration of the medicine:
| Interval | Test \(x\)-value | \(f'(x)\) | Conclusion |
|---|---|---|---|
| \(\textcolor{green}{(-∞, 1.44)}\) | \(1\) | \(f'(1)= 23\) | \(f\) is Increasing |
| \(\textcolor{red}{(1.44, ∞)}\) | \(1.5\) | \(f'(1.5)= -2\) | \(f\) is Decreasing |
We can determine that there is a maximum at \(t = 1.44\;[s]\).
We can substitute \(1.44\) for \(t\) in the original function to determine its concentration value at that time:
\(C(1.44) = 150(1.44)(0.5)^{1.44}\)
\(C(1.44) = 216(0.37)\)
\(C(1.44) = 79.61\)
Therefore, we can determine that the highest concentration value of the medicine is \(\boldsymbol{79.61\;[\textbf{parts}/\textbf{million}]}\).
ii. In order to determine how quickly the concentration is decreasing after \(2\) hours, we can substitute \(2\) into the derivative function:
\(C'(2) = 150(0.5)^2 + 150(2)(0.5)^2\cdot \ln(0.5)\)
\(C'(2) = 150(0.25) + 300(0.25)(-0.69)\)
\(C'(2) = 37.5 - 51.75\)
\(C'(2) = -14.25\)
Therefore, we can determine at \(2\) hours, the medicine is decreasing at a rate of \(\boldsymbol{-14.25\; [\textbf{parts per milllion}/\textbf{hour}]}\).
Categorizing Derivatives
Try and fit the following derivatives into their correct description.
\(\cos(5 - 3x)\)
\((6x^2 + 4)(9 - 7x^3)\)
\(\cfrac{4x - 8}{e^x}\)
\((\ln x -7)^2\)
\(\cfrac{x^2}{2 - x}\)
\((2x^2 + 5x + 3)(x - 4)\)