Introduction

Welcome to another article about derivative rules! We are going to learn how to differentiate more functions, expanding into exponential, logarithmic, and trigonometric functions. 

Derivatives Rules

$\frac{d}{dx}[\cos (x)]=-\sin (x)$

This is the first rule of our article, and it is related to this other rule.

$\frac{d}{dx}[\sin (x)]=\cos (x)$

Take note that the derivative of $\cos (x)$ is negative $\sin (x)$. Do not forget to add the negative sign.

$\frac{d}{dx}[e^x]=e^x$

Notice that this function’s derivative is itself. This is one of the only functions with this property where $f(x)$ and $f’(x)$ are the exact same. 

$\frac{d}{dx}[\ln (x)]=\frac{1}{x}$

The natural log function’s slope can be represented with simply the inverse function, how exciting! 

Take note that the domain of $\ln (x)$ is $(0,\infty )$, which means that the domain of the derivative is also $(0,\infty )$. Remember that wherever a function is discontinuous or outside its domain, it will not have a derivative. Though, you won’t really have to worry about this very much in AP Calculus when you’re computing derivatives.

$\frac{d}{dx}[a^x]=a^x\ln (a)$

This is a more general rule, where $a$ is a constant number. This general rule kind of shows why $\frac{d}{dx}[e^x]=e^x$. Remember that $\ln (e)=1$

$\frac{d}{dx}[e^x]=e^x\ln (e)=e^x$

Another rule very similar is:

$\frac{d}{dx}[{\log }_a(x)]=\frac{1}{x\ln (a)}$.

Again, $a$ is a constant and it's simply a matter of memorizing here. This rule also shows the reason why a derivative is what it is, except its with $\frac{d}{dx}[\ln (x)]=\frac{1}{x}$. Take note that $\ln (e)=1$.

$\frac{d}{dx}[{\log }_e(x)]=\frac{1}{x\ln (e)}=\frac{1}{x}$

If you’re wondering about the origins of this rule, it comes from the logarithmic change of base formula. 

Differentiating Functions

Example 1

Find $f^{\prime }(x)$ if $f(x)=3\sin (x)+\ln (x)-e^x+4$.

To differentiate this function, we simply need to apply our derivative rules. If you need to look back at them, you can scroll back up. 

$\frac{d}{dx}[3\sin (x)+\ln (x)-e^x+4]$

$=\frac{d}{dx}[3\sin (x)]+\frac{d}{dx}[\ln (x)]-\frac{d}{dx}[e^x]+\frac{d}{dx}[4]$

$=3\cos (x)+\frac{1}{x}-e^x$

Differentiating Another Function

Differentiate $f(x)={\log }_7(x)-2\sin (x)-3x^2+3\sqrt{3}$ with respect to $x$. 

It should be relatively simple to differentiate after how much practice you have, unless you haven’t been practicing all this time. 

$\frac{d}{dx}[{\log }_7(x)-2\sin (x)-3x^2+3\sqrt{3}]=\frac{d}{dx}[{\log }_7(x)]-\frac{d}{dx}[2\sin (x)]-\frac{d}{dx}[3x^2]+\frac{d}{dx}[3\sqrt{3}]$

Time to differentiate, and I won’t go through all the commentary as all you need to know to differentiate is above you.

$\frac{d}{dx}[{\log }_7(x)]-\frac{d}{dx}[2\sin (x)]-\frac{d}{dx}[3x^2]+\frac{d}{dx}[3\sqrt{3}]=\frac{1}{x\ln 7}-2\cos (x)-6x+0$

Derivative at a Point

Consider $f(x)=\ln x$. Calculate $f’(2)$, $f’(0)$, and $f’(-2)$.

We learned from above that the derivative of $\ln x$ is $\frac{1}{x}$, and so we can simply plug in the numbers.

$f’(x)=\frac{1}{x}$, so $f’(2)=\frac{1}{2}$.

$f’(0)$ is undefined ($\frac{1}{0}$).

$f’(-2)$ is actually DNE because the domain of $\ln x$ is $(0,\infty )$ and $x=-2$ is outside the domain.

Notice that even when $f’(-2)$ has a valid value being $\frac{1}{-2}$, you are not able to say that this is the derivative for the function $\ln (x)$ at $x=-2$.

Recognizing Definition of a Derivative

On the AP Calculus AB/BC exam, it is very likely that you will find a question that has the expressions and numbers plugged into the definition of the derivative. These types of questions will ask you to find the limit, or calculate the value or expression. 

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

If a function is chosen but no number is plugged in (only evaluating for $f(x)$, the general function), then you will see the $x$ variable in the definition. For example, if I were to plug in $f(x)=\sin (x)$, this is what the definition would look like.

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{\sin (x+h)-\sin (x)}{h}$

It is very important that you correlate $f(x+h)$ with $\sin (x+h)$, and $f(x)$ with $\sin (x)$. Doing this, especially going in reverse (figuring out the derivative with the limit definition of the derivative with all the values plugged in) will allow you to figure out things much more easily.

With $f(x)$ plugged in, I can also plug in the $x$-value at which I want the derivative at. If I wanted the slope of $f(x)$ at $x=\pi$, the following expression would give me the same value as $f’(\pi )$

$f^{\prime }(\pi )=\underset{h\to 0}{\lim }\frac{\sin (\pi +h)-\sin (\pi )}{h}=\underset{h\to 0}{\lim }\frac{\sin (\pi +h)-0}{h}=\underset{h\to 0}{\lim }\frac{\sin (\pi +h)}{h}$

In the following problem, I would like for us to do the reverse. Recognize the function and numbers that are being plugged in. Remember that you do not need to figure out if the definition is attempting to solve the derivative at a point if the $x$-variable is present, as this means a value for $x$ has not been chosen.

Differentiating Through the Definition of the Limit

Evaluate the expression $\underset{h\to 0}{\lim }\frac{e^h-1}{h}$.

We haven’t discussed another way to solve this derivative, which is to straight up solve the limit as it is. However, this is ineffective unless you know certain formulas and patterns that are not needed for the AP exam. Pattern recognition and conceptual understanding is much more important.

Notice how the $x$-variable is not present. This means that this is the derivative of a function at a point. If we correlate each function to its corresponding component in the definition, this means that:

$f(x)=1$ and $f(x+h)=e^h$

We will need to find a value of $x$ and a function that both satisfies this. It is likely that you will need to try out certain formulas, but starting off with something along the lines of $e^x$ is reasonable in this case. 

If we set $f(x)=e^x$, then we get:

$e^x=1$ and $e^{x+h}=e^h$

The only $x$-value that satisfies $e^x=1$ is $x=0$, and plugging in $x=0$ into $e^{x+h}=e^h$ also yields a true statement. From this info, we can conclude that we are trying to find the derivative of $f(x)=e^x$ at $x=0$, or in other words, $f’(0)$. 

Now that we know the original function, we can simply differentiate and plug. If you remember, the derivative of $e^x$ is $e^x$, and so $f’(x)=e^x$. Plug in $x=0$, and you get $f’(0)=e^0=1$. The answer to the question is a solid $1$.

Closing Remarks

Because differentiation is so important, and it is also important to memorize these rules, there will be more practice questions with this article. I would like to tell you that throughout Calculus, you will use these rules the most frequently.

Practice