When calculating the derivatives of polynomials (especially more complex ones), it might get a little tedious using First Principles to determine every one. As a result, there are a set of rules used to make this process more straightforward:
| 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 | \(f(x) = g(x) - H(x)\) | \(f'(x) = g'(x) - H'(x)\) |
Make sure to use the calculator here in order to verify your answers for the Power Rule!
Differentiate the following functions. Simplify first if necessary:
i. We can take the derivative of each individual term using the Sum Rule. We can also differentiate the first 2 terms (\(5x^6\) and \(-4x^3\)) using the Power Rule. Additionally, we can differentiate the last term using the Constant Rule:
\(y' = \textcolor{red}{3}(4x^{\textcolor{red}{3}-1}) + 0\)
\(y' = 30x^5 -12x^2\)
Therefore, we can determine that \(\boldsymbol{y' = 30x^5 -12x^2}\).
ii. First, we can rewrite the function as such:
\(f(x) = -3x^5 + 8x^{\frac{1}{2}} - 9x^{-1}\)
Next, we can take the derivative of each individual term using the Sum/Difference Rules. We can also differentiate each term by using Power Rule:
\(f'(x) = \textcolor{red}{5}(-3x^{\textcolor{red}{5}-1}) + \cfrac{\textcolor{red}{1}}{\textcolor{red}{2}}(8x^{\textcolor{red}{\frac{1}{2}}-1}) - (\textcolor{red}{-1})(9x^{\textcolor{red}{-1}-1})\)
\(f'(x) = -15x^4 + 4x^{-\frac{1}{2}} + 9x^{-2}\)
Finally, we can express the derivative in simplest form by rewriting it with positive exponents:
\(f'(x) = -15x^4 + \cfrac{4}{\sqrt{x}} + \cfrac{9}{x^2}\)
Therefore, we can determine that \(\boldsymbol{f'(x) = -15x^4 + \cfrac{4}{\sqrt{x}} + \cfrac{9}{x^2}}\).
\(g(x) = (2x-3)(x+1)\)
First, we can create an expanded polynomial:
\(g(x) = (2x)(x) + (-3)(x) + (2x)(1) + (-3)(1)\)
\(g(x) = 2x^2 -3x + 2x - 3\)
\(g(x) = 2x^2 -x - 3\)
Next, we can differentiate each individual term using the Difference Rule. Additionally, we can differentiate the first 2 terms (\(2x^2\) and \(-x\)) using Power Rule. Lastly, we can differentiate the last term (\(3\)) using Constant Rule:
\(g'(x) = \textcolor{red}{2}(2x^{\textcolor{red}{2}-1}) -\textcolor{red}{1}(x^{\textcolor{red}{1}-1}) - \textcolor{red}{0}(3)\)
\(g'(x) = 4x -1\)
Therefore, we can determine that \(\boldsymbol{g'(x) = 4x -1}\).
\(h(x) = \cfrac{-8x^6+8x^2}{4x^5}\)
First, we can rewrite the function as such:
\(h(x) = \cfrac{-8x^6}{4x^5} + \cfrac{8x^2}{4x^5}\)
We can now simplify the function:
\(h(x) = -2\cfrac{x^{\cancel{6}1}}{\cancel{x^5}} + 2\cfrac{\cancel{x^2}}{x^{\cancel{5}3}}\)
\(h(x) = -2x + 2x^{-3}\)
Next, we can take the derivative of each individual term using the Sum/Difference Rules. We can also differentiate the function by using Power Rule on both terms:
\(h'(x) = \textcolor{red}{1}(-2x^{\textcolor{red}{1}-1}) + (\textcolor{red}{-3}(2x^{\textcolor{red}{-3}-1}))\)
\(h'(x) = -2 -6x^{-4}\)
Finally, we can express the derivative in simplest form by rewriting it with positive exponents:
\(h'(x) = -2 -\cfrac{6}{x^4}\)
Therefore, we can determine that \(\boldsymbol{h'(x) = -2 - \cfrac{6}{x^4}}\).