A derivative is commonly used in calculus to determine the slope of the function's tangent at any point. A function's First Principles refers to calculating the derivative from the definition that uses the limit.
The differentiation operation, used to determine the derivative, is algebraically expressed as:
\(f'(x) = \cfrac{df}{dx} = \lim\limits_{h\to0}\cfrac{\textcolor{red}{f(x+h)}-\textcolor{blue}{f(x)}}{h}\)
In Prime Notation, the derivative is expressed as:
\(f'(x)\)
OR
\(y'\)
In Leibniz Notation, the derivative is expressed as:
\(\cfrac{dy}{dx}\)
OR
\(\lim\limits_{h\to0}\cfrac{Δy}{Δx}\)
This translates to the derivative of \(y\) with respect to \(x\).
NOTE: Leibniz Notation can use different variables aside from just \(x\). This is important to keep in mind when dealing with functions with more than one independent variable!
Example
Determine the derivatives of the following functions:
- \(f(x) = x^4\)
- \(g(x) = \cfrac{1}{x^2}\)
i. First, we can plug \(x^4\) into the First Principles formula:
\(f'(x) = \lim\limits_{h\to0} \left[\cfrac{\textcolor{red}{f(x+h)}-\textcolor{blue}{f(x)}}{h}\right]\)
\(f'(x) = \lim\limits_{h\to0} \left[\cfrac{\textcolor{red}{(x+h)^4}-\textcolor{blue}{x^4}}{h}\right]\)
Next, we can simplify the equation by expanding the numerator and collecting like terms:
\(f'(x) = \lim\limits_{h\to0} \left[\cfrac{(x+h)^4-x^4}{h}\right]\)
\(f'(x) = \lim\limits_{h\to0} \left[\cfrac{x^4+4x^3h+6x^2h^2 +4xh^3+h^4-x^4}{h}\right]\)
\(f'(x) = \lim\limits_{h\to0} \left[\cfrac{4x^3h+6x^2h^2 +4xh^3+h^4}{h}\right]\)
Then, we can take a common factor out of the numerator:
\(f'(x) = \lim\limits_{h\to0} \left[\cfrac{\cancel{h}(4x^3+6x^2h +4xh^2+h^3)}{\cancel{h}}\right]\)
\(f'(x) = \lim\limits_{h\to0}[4x^3+6x^2h +4xh^2+h^3]\)
Finally, we can substitute \(0\) for \(h\) and simplify to determine the final derivative:
\(f'(x) = 4x^3+(6x^2)(\textcolor{red}{0}) + (4x)(\textcolor{red}{0})^2 + (\textcolor{red}{0})h^3\)
\(f'(x) = 4x^3\)
Therefore, we can determine that \(f'(x) = 4x^3\).
ii. First, we can plug \(\cfrac{1}{x^2}\) into the First Principles formula:
\(g'(x) = \lim\limits_{h\to0} \left[\cfrac{\textcolor{red}{f(x+h)}-\textcolor{blue}{f(x)}}{h}\right]\)
\(g'(x) = \lim\limits_{h\to0} \left[\left(\textcolor{red}{\cfrac{1}{(x+h)^2}} - \textcolor{blue}{\cfrac{1}{x^2}}\right)\left(\cfrac{1}{h}\right)\right]\)
Next, we can cross multiply the terms in the main expression to place all the terms under a common denominator.
\(g'(x) = \lim\limits_{h\to0}\left[\left[\left(\cfrac{\textcolor{blue}{x^2}}{\textcolor{blue}{x^2}}\right) \left(\cfrac{1}{(x+h)^2}\right) - \left(\cfrac{\textcolor{red}{(x+h)^2}}{\textcolor{red}{(x+h)^2}}\right) \left(\cfrac{1}{x^2}\right)\right]\cdot \left[\cfrac{1}{h}\right]\right]\)
\(g'(x) = \lim\limits_{h\to0} \left[\left(\cfrac{x^2}{x^2(x+h)^2} - \cfrac{(x+h)^2}{x^2(x+h)^2}\right)\left(\cfrac{1}{h}\right)\right]\)
Then, we can multiply the main expression by \(\cfrac{1}{h}\):
\(g'(x) = \lim\limits_{h\to0} \left[\cfrac{x^2 - (x+h)^2}{hx^2(x+h)^2}\right]\)
After, we can expand the second term in the numerator:
\(g'(x) = \lim\limits_{h\to0} \left[\cfrac{x^2 - (x+h)^2}{hx^2(x+h)^2}\right]\)
\(g'(x) = \lim\limits_{h\to0} \left[\cfrac{x^2 - (x+h)(x+h)}{hx^2(x+h)^2}\right]\)
\(g'(x) = \lim\limits_{h\to0} \left[\cfrac{x^2 - (x^2 + 2xh + h^2)}{hx^2(x+h)^2}\right]\)
Subsequently, we can simplify the equation by expanding the numerator and collecting and cancelling like terms:
\(g'(x) = \lim\limits_{h\to0} \left[\cfrac{x^2 - x^2 - 2xh - h^2}{hx^2(x+h)^2}\right]\)
\(g'(x) = \lim\limits_{h\to0} \left[\cfrac{\cancel{h}(-2x -h)}{\cancel{h}x^2(x+h)^2}\right]\)
\(g'(x) = \lim\limits_{h\to0} \left[\cfrac{-2x -h}{x^2(x+h)^2}\right]\)
Finally, we can substitute \(0\) for \(h\) and simplify to find the final derivative:
\(g'(x) = \cfrac{-2x - \textcolor{red}{0}}{x^2(x+\textcolor{red}{0})^2}\)
\(g'(x) = \cfrac{-2\cancel{x}}{x^2\cdot x^{\cancel{2}}}\)
\(g'(x) = \cfrac{-2}{x^3}\)
Therefore, we can determine that \(g'(x) = \cfrac{-2}{x^3}\).
For the function, \(h(x) = x^2\):
- Find the derivative using 1st Principles.
- Use the derivative to calculate the slopes at \(x=-3,0,1\).
- Graph the function and the tangents found above to show the graphical representation of the derivative.
Show Answer
i. First, we can plug \(x^2\) into the First Principles formula:
\(h'(x) = \lim\limits_{h\to0} \left[\cfrac{\textcolor{red}{f(x+h)}-\textcolor{blue}{f(x)}}{h}\right]\)
\(h'(x) = \lim\limits_{h\to0} \left[\cfrac{\textcolor{red}{(x+h)^2}-\textcolor{blue}{x^2}}{h}\right]\)
Next, we can simplify the equation by expanding the numerator and collecting and cancelling like terms:
\(h'(x) = \lim\limits_{h\to0} \left[\cfrac{(x+h)^2-x^2}{h}\right]\)
\(h'(x) = \lim\limits_{h\to0} \left[\cfrac{x^2 + 2xh + h^2-x^2}{h}\right]\)
\(h'(x) = \lim\limits_{h\to0} \left[\cfrac{\cancel{h}(2x+h)}{\cancel{h}}\right]\)
\(h'(x) = \lim\limits_{h\to0}[2x+h]\)
Finally, we can substitute \(0\) for \(h\) to get our derivative:
\(h'(x) = 2x+\textcolor{red}{0}\)
\(h'(x) = 2x\)
Therefore, we can determine that \(h'(x)=2x\).
ii. We can substitute the respective \(x\)-values for \(x\) to determine the slopes at those particular points:
\(h(\textcolor{red}{-3}) = 2(\textcolor{red}{-3})\)
\(h(\textcolor{red}{-3}) = -6\)
\(h(\textcolor{green}{0}) = 2(\textcolor{green}{0})\)
\(h(\textcolor{green}{0}) = 0\)
\(h(\textcolor{blue}{1}) = 2(\textcolor{blue}{1})\)
\(h(\textcolor{blue}{1}) = 2\)
Therefore, we can determine that slope at \(x=-3\) is \(-6\), the slope at \(x=0\) is \(0\), and the slope at \(x=1\) is \(2\).
iii. Given that we have found our slope values, we are now able to graph our function with the pertinent information. Since we are graphing a quadratic function, we are able to find the points fairly easily:
Sketching Derivatives of A Function
When evaluating a graph of a function, there a few important things to take into consideration. These will help us sketch its derivative:
Original Function |
Derivative Function |
\(f(x)\) is increasing (positive slope) |
\(f'(x)\) is above the x-axis |
\(f(x)\) is decreasing (negative slope) |
\(f'(x)\) is below the x-axis |
\(f(x)\) has a Turning Point |
\(f'(x)\) is \(0\) |
As an example, we can look at a Position-Time graph to determine how its different portions affect its derivative:
From the graph above, we can determine that the horizontal lines represent the portions of the function that have a slope of \(0\). The lines moving upward and downward represent portions of the function containing constant slopes that we can calculate.
Below is how the function would get represented as a Velocity-Time graph. This also represents the derivative of the original function:
As an additional example, we can look at a Qudratic Function and how it affects its derivative, a Linear Function:
While the Quadratic Function is increasing, it's getting gradually less steep until it reaches its Turning Point. While it's decreasing, it's gradually getting steeper.
The Linear Function begins above the \(x-\)-axis decreases at a constant rate until \(x = 0\). It then continues to decrease at a constant rate below the \(x\)-axis.
Enter in a coefficient for quadratic monomial or click on the New Question button for a random value. Then, use the First Principles formula to solve for the derivative. This will also create a graph comparing the original function to the derivative.