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Derivative Examples

As we continue to work through derivatives, we are now ready to apply our derivative rules to different examples. If you need a review of the different derivative rules please visit the posts Power Rule, Products, Quotient Rule, and Chain Rule.

Simplify and Differentiate

In our first example we will find the derivative of

As we begin to take the derivative, we will want to simplify the expression so that we can make things easier on ourselves. We begin by simplifying the function and getting,
\frac{x^{2}-2}{x}+3e^{x}&=\frac{1}{x}(x^{2}-2)+3e^{x} \\
&=\frac{1}{x}x^{2}-2\frac{1}{x}+3e^{x} \\

Now that we have simplified the expression, we are now ready to take the derivative. We will continue by distributing the derivative across sum and differences and factoring out constants. We now have
\left(\frac{x^{2}-2}{x}+3e^{x}\right)’&=\left(x-\frac{2}{x}+3e^{x}\right)’ \\
&=(x)’-2(x^{-1})’+3(e^{x})’ \\
&=1-2(-x^{-2})+3e^{x} \\
Here, we calculated our derivatives using the power rule after we had distributed the derivative.

Product Rule

For our next example, we won’t be able to simplify the problem enough in order to use the power rule alone. Instead we will have to find another way to take the derivative. We will now find

In this case, we want to look at the function from a broad perspective. The large picture that we see here is that there is a product of two functions. Because of this, we will want to use the product rule. Before doing so, we recall that \((SF)’=SF’+FS’\), so we find

  • \(F=x^{3}+2\),
  • \(S=\ln(x)\).
  • \(F’=3x^{2}\)
  • \(S’=\frac{1}{x}\)

Combining this, we have
\frac{d}{dx}[(x^{3}+2)\ln(x)]&=SF’+FS’ \\

Note that since we took this one step at a time and found our derivatives individually, we were able to find to the derivative of the more complicated function. As we continue, we will want to take things in as small of steps as possible.

Multiple rules

We will now find

Quotient Rule

As we did in the last example, we begin by looking at the problem with a broad perspective. The first thing I see when I look at this problem is the quotient. Note that, you may see the \(\sin(x)\) as it is the more intimidating part of the problem; however, I want to focus on the largest part of the part, not the scariest. Therefore, I want to completely ignore details until later, so the big line going across the example would be the first visible element.

Because we have quotient, we will need the quotient rule. Recall then that this is
We now see that

  • \(T=(x^{2}+2x)(\sin(x)+5)\),
  • \(B=3x+2\),
  • \(T’= ?\),
  • \(B’=3\).

Note that I left the derivative of the top as a question mark. I did this because I wasn’t immediately able to find the derivative. That is, if I have to think about it at all, I will break this portion off as a separate problem. The reason to do this is, again, because the smaller you make each step, the less likely you are to make mistakes. We, therefore, forget the original problem and focus on the smaller problem.

Product Rule

In order to find
we again look at the broad view of the problem. In this case, we see that there is a product of functions. We will then use the product rule, so we remind ourselves of that. The product rule is
We now find that

  • \(F=x^{2}+2x\),
  • \(S=\sin(x)+5\),
  • \(F’=2x+2\),
  • \(S’=\cos(x)\).

As we combine this, we find that
\frac{d}{dx}&(x^{2}+2x)(\sin(x)+5) \\
&=SF’+FS’ \\

Finishing the problem

Now that we’ve found the derivative of the top as a separate problem, we now have to combine the answer together with the original problem. In order to remind us of what we found, we have that

  • \(T=(x^{2}+2x)(\sin(x)+5) \),
  • \(B=3x+2\),
  • \(T’=(\sin(x)+5)(2x+2)+(x^{2}+2x)(\cos(x))\),
  • \(B’=3\).

We, therefore, get that
\frac{d}{dx}&\left(\frac{(x^{2}+2x)(\sin(x)+5)}{(3x+2)}\right) \\
&=\frac{BT’-TB’}{B^{2}} \\

I should note here that, since we have the derivative and won’t be using it, I wouldn’t bother to simplify the derivative at all. Since we have the answer, the extra time required to simplify the problem could be spent working on other examples.

Chain Rule

As our final example for this post, we will find

As we look at this example, the broadest part of the problem we see is that there is a function inside of another function. We will have to use the chain rule. Again, the chain rule is
where \(o\) is the outside function and \(i\) is the inside function. In this case, we will have

  • \(o(i)=\tan^{-1}(i)\),
  • \(i(x)=x^{4}+2x\),
  • \(o'(i)=\frac{1}{1+i^{2}}\),
  • \(i'(x)=4x^{3}+2\).

If we now combine this together we get that
[\tan^{-1}(x^{4}+2x)]’&=o'(i)*i'(x) \\
&=\frac{1}{1+i^{2}}(4x^{3}+2) \\

In the last step, we just substituted back in the inside function to arrive at a function involving only \(x\)s.


In each of these cases, we had to begin by trying to take in the problem and see what the best approach we should use for the problem. As we worked on this, we made sure not to do too much at the same time so that we didn’t make any calculation mistakes. While we have examples that go through each of the derivative rules we learned, there are many other ways that we can combine these rules to come up with even more examples. As such, I hope you can use these to see the process needed to work on derivatives of general.

I hope you learned something today, and that you enjoyed the read while you were at it. If this was helpful, make sure to like the post and share it on Social Media with anyone else that may find it useful.

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