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Introduction
I have chosen to discuss the properties of the derivative that you’ll use in finding derivatives of more complex functions at this point in the notes (after using the power rule, but before going on to more complicated examples) because I think it fits best within the curriculum at this point. Your class’ order may differ slightly. This will make working with these properties easier than if I had introduced them before you learned the power rule.
Important Properties of the Derivative
Here are multiple important properties and results for the derivative that will be necessary to invoke henceforth:
Properties 1 and 2 pertain to the linearity of the derivative; all operators that are “linear” adhere to these principles. Property 3 was shown in the power rule section. Property 4 is a result of using the power rule on  (which itself is 1, the derivative of which returns 0) coupled with property 1 to remove
 (which itself is 1, the derivative of which returns 0) coupled with property 1 to remove  from the derivative and leave simply 1 inside.
 from the derivative and leave simply 1 inside.
My full notes on this section provide detailed proofs of each of these properties (using the definition of the derivative and/or the power rule) as well as more discussion on their origins, usage, and relevant examples.
Current Examples
Here are some examples that utilize these properties, involving derivatives we’ve already seen so far. If you’ve already covered later types of derivatives in this class, you’ll also want to check out the more advanced examples below these.
Example 1
Let  . Find
. Find 
 
Although an irrational one, this function is still a simple constant, so its derivative is null:
      ![Rendered by QuickLaTeX.com \[ \diff{}{x} \Big[29 \pi \Big] = 0 \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-2ad74dc63c297c6e6812c9617d85353c_l3.png)
Example 2
Let  . Find
. Find 
This is really the application of two “properties”, first the constant property (because of linearity), and second the fact that the derivative of a variable with respect to itself is 1. Therefore,
      ![Rendered by QuickLaTeX.com \[ \diff{}{x} \Big[ -22x \Big] = -22 \diff{}{x} \Big[x \Big] = -22 \Big[ 1 \Big] = -22 \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-bbfd5de6e538bd04447eaf1adde4df18_l3.png)
Example 3
Find the derivative of 
Because these two terms are added together, if you can find the derivative of each of them separately, you can find the complete derivative:
      ![Rendered by QuickLaTeX.com \[ \diff{h}{x} = \diff{}{x} \Big[x^3 \Big] - \diff{}{x} \Big[x^2 \Big] = 3x^2 - 2x \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-6871c856b4a9608f9744ea190c1d47be_l3.png)
Just as the original function was a sum, so is its derivative (provided nothing cancels).
Example 4
Differentiate 
Using the derivative properties coupled with the power rule produces
      ![Rendered by QuickLaTeX.com \[ z'(x) = -3 \Big[ 7x^6 \Big] + 4 \Big[ 6x^5 \Big] + 4 \Big[ \frac{5}{4} x^{1/4} \Big] - 1 + 0 \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-faf23ff551d35ce06c16fb46435a7331_l3.png)
      ![Rendered by QuickLaTeX.com \[ = -21x^6 + 24x^5 + 5x^{1/4} - 1 \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-efc74256a0c6c0e9f26f8b0bf1454ec4_l3.png)
Example 5
Compute the derivative of  and then find what it equals when x=4. This is called “evaluating” the derivative.
 and then find what it equals when x=4. This is called “evaluating” the derivative.
We first proceed with the calculation of the derivative itself in the normal way, and then we worry about evaluating it. Evaluation of the derivative always comes after its general form is found first.
      ![Rendered by QuickLaTeX.com \[ y'(x) = 6x^2 + \frac{3}{2}x^{1/2} - 6 \cdot \frac{1}{2}x^{-1/2} + 2x + 1 + 0 \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-37272e2f819c1a234610ab69f6f0e66b_l3.png)
      ![Rendered by QuickLaTeX.com \[ = 6x^2 + \frac{3\sqrt{x}}{2} - \frac{3}{\sqrt{x}} + 2x + 1 \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-472e02950436334b627cee8019e49c73_l3.png)
We can now “evaluate” it at  by plugging it right into our result:
 by plugging it right into our result:
      ![Rendered by QuickLaTeX.com \[ y'(4) = 6 \cdot 4^2 + \frac{3\sqrt{4}}{2} - \frac{3}{\sqrt{4}} + 2(4) + 1 \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-a7dc9462ad85602ecde4e78580dda5ca_l3.png)
      ![Rendered by QuickLaTeX.com \[ = 96 + 3 - \frac{3}{2} + 8  + 1= 106.5 \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-0b304994407b5f17389746dde927cd98_l3.png)
The notation  is pronounced “y prime of four” and means to find the derivative and then evaluate it at 4. This concept will persist throughout this class, and you can find more such examples in my full notes on this section.
 is pronounced “y prime of four” and means to find the derivative and then evaluate it at 4. This concept will persist throughout this class, and you can find more such examples in my full notes on this section.
I have only included a handful of such examples in this section. However, you’ll continue working on dervatives of power functions using these properties in the product rule, quotient rule, and chain rule sections.
Later Practice with Derivative Properties
The properties above apply when taking derivatives of any functions, so here are a few examples using derivative types from the next chapter on advanced derivative methods. You can neglect these examples for now if you aren’t past this current section in your class yet.
Example 6
Find the derivative of 
We can apply the linearity of the derivative in the same way as with the other types of functions, provided we know the individual trigonometric derivatives. That goes as follows:
      ![Rendered by QuickLaTeX.com \[ y' = -\sin x + 3 \cos x - 12 \sec^2 x \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-db6323a0e443519f3eefe31c91dfb86d_l3.png)
Example 7
Let  . Find
. Find 
We can apply the linearity of the derivative in the same way as with the other types of functions, provided we know the individual exponential derivatives. The derivative of an exponential function is the same function, multiplied by the derivative of the exponent, so, pulling constants out as well, that goes as follows:
      ![Rendered by QuickLaTeX.com \[ J'(x)= e^x + 2e^{x^2} \cdot 2x + 3e^{x^3} \cdot 3x^2 + 0 \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-aaaa50d858c583aa095e00f988fa58c3_l3.png)
      ![Rendered by QuickLaTeX.com \[ = e^x + 4xe^{x^2} + 9x^2 e^{x^3} \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-76ec48b0dbae2d1f87a7d819b39a7136_l3.png)
Notice that, although irrational,  is still just a constant, so its derivative is 0.
 is still just a constant, so its derivative is 0.
Example 8
Find  where
 where  
 
In addition to the trig function, I’ve also thrown in some logarithms and inverse trig functions for which to take the derivatives. The properties from this section still apply.
      ![Rendered by QuickLaTeX.com \[   f'(x) = 12 \cdot -\csc x \cot x + \frac{1}{x^2 + 1} \cdot (2x) - \frac{1}{x^2 + 1} + \frac{1}{\arcsin x} \cdot \frac{1}{\sqrt{1-x^2}} \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-296d7d98f28c4ba1b5dc028c28e1412c_l3.png)
      ![Rendered by QuickLaTeX.com \[ = -12\csc x \cot x + \frac{2x-1}{x^2+1} + \frac{1}{\arcsin x \sqrt{1-x^2}} \]](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-5ac48350b4430fa8ce50519486e9e5ec_l3.png)
where in the last line, I added the two fractions that had the same denominator and combined them.
Need more? Here you can get the full version of these notes!
![Rendered by QuickLaTeX.com \begin{equation*} \diff{}{x} \Big[ c f(x) \Big] = c \diff{}{x} \Big[f(x) \Big]  \end{equation*}](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-4b16fa7561c3686ef4ae77ae60477bcc_l3.png)
![Rendered by QuickLaTeX.com \begin{equation*} \diff{}{x} \Big[ f(x) \pm g(x) \Big] = \diff{}{x} \Big[f(x) \Big] \pm \diff{}{x} \Big[g(x) \Big]  \end{equation*}](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-e136cc5f46bd1a8b1c2e6461732be239_l3.png)
![Rendered by QuickLaTeX.com \begin{equation*} \diff{}{x} \Big[ x \Big] = 1  \end{equation*}](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-da571732c84a730c67f2ac29efe700bd_l3.png)
![Rendered by QuickLaTeX.com \begin{equation*} \diff{}{x} \Big[ c \Big] = 0  \end{equation*}](https://www.mathhelpandtutoring.com/wp-content/ql-cache/quicklatex.com-9e647004f2dcb2771c530d872e4473a2_l3.png)