Basic Differentiation Rules: Constant, Power, Product, Quotient and Trig Rules

This section covers:

Note that you can use www.wolframalpha.com (or use app on smartphone) to check derivatives by typing in “derivative of x^2(x^2+1)”, for example. Really cool!

I promised you that I’d give you easier way to take derivatives, and the constant, power, product, quotient and basic trigonometry function rules make it much, much easier.

Let’s first talk about some notation. When we take the derivative, say of the function  \(f\left( x \right)=x+3\), we typically say that we are taking the derivative of \(\boldsymbol {y=f\left( x \right)}\) with respect to \(\boldsymbol {x}\), or whatever the independent variable is (later we’ll take the derivative with respect to more than one variable). When we take the derivative with respect to \(x\), we may see it written the following ways: \(\displaystyle {f}’\left( x \right),\,\,{y}’,\,\,\frac{{dy}}{{dx}},\,\,\frac{{d\left( {f\left( x \right)} \right)}}{{dx}},\,\,\frac{d}{{dx}}\left( {f\left( x \right)} \right)\)  (and you may see it differently, such as  \(\displaystyle \frac{{df}}{{dx}},\,\frac{d}{{dx}}y,\,{{D}_{x}}y\)). It all basically means the same thing.

Note that when you have an algebraic expression that’s a sum or difference (like \({{x}^{2}}+x+3\)), you can separate the expressions and take the derivative of each part (like \({{x}^{2}}\)  and \(x\) and 3) and add (or subtract) them together! This is called the Sum and Difference Rule.

Note that there are examples for all these rules here.

(For really complicated functions that appear to need power, product and/or quotient rules, it may be easier to take the “ln” (natural logarithm) of each side to differentiate, like we did here.)

Constant Rule

This is an easy one; whenever we have a constant (a number by itself without a variable), the derivative is just 0. For example, if we have \(y=3\) and want the derivative of that function, it’s just 0.

Here is what it looks like in Theorem form:

If \(c\) is a constant real number, then

\(\displaystyle \frac{d}{{dx}}\left( c \right)=0\)

 

When you think about this, it makes sense; remember that the graph of a constant, like \(y=3\), is just a horizontal line, and what’s the slope of a horizontal line? 0!

Power Rule

The way I remember the power rule is take the exponent of a function and move it to the front (to multiply the rest by, including any coefficients), and then take the exponent down a level. In order to use this though, we have to make sure it’s only \(\boldsymbol {x}\) (or whatever the variable is) that’s raised to that exponent.

For example, if we have \(y=4{{x}^{3}}\), we move the 3 in front, and bring the \(x\) cubed down to \(x\) squared: \(\displaystyle \frac{{dy}}{{dx}}=3\cdot 4{{x}^{{3-1}}}=12{{x}^{2}}\). Interestingly enough, when we take the derivative of the volume of a sphere with respect to its radius, \(\displaystyle V=\frac{4}{3}\pi {{r}^{3}}\), we get the surface area of a sphere, \(\displaystyle \frac{{dV}}{{dr}}=3\cdot \frac{4}{3}\pi {{r}^{{3-1}}}=4\pi {{r}^{2}}\). Not a coincidence!

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Here is what it looks like in Theorem form:

If \(n\) is a rational number, then

\(\displaystyle \frac{d}{{dx}}\left( {{{x}^{n}}} \right)=n{{x}^{{n-1}}}\)

 

We have to be a little careful here, since, for \(f\) to be differentiable at \(x=0\), for example, \({{x}^{{n-1}}}\) must be defined for when \(x=0\). For example, \(\displaystyle \sqrt{x}={{x}^{{\frac{1}{2}}}}\) is defined at \(x=0\), but not differentiable at \(x=0\), since \({{x}^{{-\frac{1}{2}}}}\) is not defined when \(x=0\).

Product Rule

The product rule gets a little more complicated, but after a while, you’ll be doing it in your sleep. Make it into a little song, and it becomes much easier. And notice that typically you have to use the constant and power rules for the individual expressions when you are using the product rule.

Only use the product rule if there is some sort of variable in both expressions that you’re multiplying. For example, use it when you have something like \({{x}^{2}}\left( {x+3} \right)\), but not something like \(\left( {5{{x}^{2}}} \right)\left( 2 \right)\); turn this into \(10{{x}^{2}}\).

Also, if you can, you can turn exponents on the bottom to negative exponents; for example, \(\displaystyle \frac{5}{{{{x}^{2}}}}=5{{x}^{{-2}}}\) .

Try to simplify your function first.

And note that when you have the product of two expressions with variables in it, the derivative is not just the product of their derivatives.

Here’s how I like to remember it: First times the derivative of the second PLUS second times derivative of the first. (Yes, that is a PLUS in the middle).

Here is what it looks like in Theorem form:

The product of two differentiable functions is differentiable and is:

\(\displaystyle \frac{d}{{dx}}\left( {f\left( x \right)\cdot g\left( x \right)} \right)=f\left( x \right){g}’\left( x \right)+g\left( x \right){f}’\left( x \right)\,\)

 

Note that there are other ways to write this such as:

\(\displaystyle \begin{align}\frac{d}{{dx}}\left( {f\left( x \right)\cdot g\left( x \right)} \right)&=f\left( x \right){g}’\left( x \right)+{f}’\left( x \right)g\left( x \right)\,\,\,\,\,\,\,\text{or}\\\frac{d}{{dx}}\left( {f\left( x \right)\cdot g\left( x \right)} \right)&={f}’\left( x \right)g\left( x \right)+f\left( x \right){g}’\left( x \right)\end{align}\)

Note that if you have a coefficient in front of two factors, you can either lump the coefficient with one of the factors (like the first one), or take it out and multiply the whole derivative later. For example, for \(y=5x{{\left( {x+1} \right)}^{3}}\), the derivative can be obtained this way:

\(\begin{align}5x\cdot 3{{\left( {x+1} \right)}^{2}}+{{\left( {x+1} \right)}^{3}}\cdot 5&=15x{{\left( {x+1} \right)}^{2}}+5{{\left( {x+1} \right)}^{3}}\\&={{\left( {x+1} \right)}^{2}}\left[ {15x+5\left( {x+1} \right)} \right]={{\left( {x+1} \right)}^{2}}\left( {20x+5} \right)\end{align}\)

or this way:

\(\begin{align}5\left[ {x\cdot 3{{{\left( {x+1} \right)}}^{2}}+{{{\left( {x+1} \right)}}^{3}}\cdot 1} \right]&=5\left[ {3x{{{\left( {x+1} \right)}}^{2}}+{{{\left( {x+1} \right)}}^{3}}} \right]\\&=5\left[ {{{{\left( {x+1} \right)}}^{2}}\left( {3x+x+1} \right)} \right]={{\left( {x+1} \right)}^{2}}\left( {20x+5} \right)\end{align}\)

Note how we took out a greatest common factor (GCF) after taking the derivative, in order to simplify the expression.

Quotient Rule

First of all, remember that you don’t need to use the quotient rule if there are just numbers on the bottom – only if there are variables on the bottom (in the denominator)!

One thing to remember about the quotient rule is to always start with the bottom, and then it will be easier. I remember it this way: Bottom times the derivative of the top minus top times the derivative of the bottom, all over the bottom squared. Note that that the top has a minus in it, not a plus.

Note that if you can separate a quotient into individual terms, it’s best to avoid the quotient rule. For example, if you have the function \(\displaystyle f\left( x \right)=\frac{{{{x}^{3}}+3}}{x}\), it’s best to divide both terms on the top by \(x\) to get  \(\displaystyle f\left( x \right)={{x}^{2}}+\frac{3}{x}={{x}^{2}}+3{{x}^{{-1}}}\) (You can’t do this with a function like \(\displaystyle f\left( x \right)=\frac{x}{{{{x}^{3}}+3}}\)).

Also note that you can typically turn any function that looks like a quotient into a product, using negative exponents. Some prefer to never use the quotient rule, but to use the product rule instead! But as you practice differentiating, you’ll find the method that works best for you.

Here is what it looks like in Theorem form:

The quotient of two differentiable functions is differentiable and is:

\(\displaystyle \frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)=\frac{{g\left( x \right){f}’\left( x \right)-f\left( x \right){g}’\left( x \right)}}{{{{{\left( {g\left( x \right)} \right)}}^{2}}}};\,\,\,\,g\left( x \right)\ne 0\)

 

Note that some people to use the Hi De Ho way to remember (Ho dHi minus Hi dHo, all over HO HO):

\(\displaystyle \frac{d}{{dx}}\left( {\frac{{HI}}{{HO}}} \right)=\frac{{HO\cdot dHI-HI\cdot dHO}}{{HO\cdot HO}}\)

List of Rules

Here are all the rules, with some examples. Do you see how with the product and quotient rules, we may need to use the constant and power rules? Also, when we have a (non-variable) coefficient, it’s typically easier to take it out first before we do the differentiation.

Note in all these cases, with what we’ve learned so far with these rules, the coefficient of the \(\boldsymbol {x}\) must be 1 (unless we can take out the coefficient from the whole expression).

Derivative Rule Definition Example
Constant Rule \(\displaystyle \frac{d}{{dx}}\left( c \right)=0\) \(\displaystyle \begin{array}{c}\color{#800000}{{f\left( x \right)=5}}\\{f}’\left( x \right)=0\end{array}\)
Power Rule \(\displaystyle \frac{d}{{dx}}\left( {{{x}^{n}}} \right)=n{{x}^{{n-1}}}\) \(\displaystyle \begin{array}{c}\color{#800000}{{f\left( x \right)=5{{x}^{9}}}}\\{f}’\left( x \right)=5\cdot 9{{x}^{{9-1}}}=45{{x}^{8}}\end{array}\)
Product Rule \(\displaystyle \frac{d}{{dx}}\left( {f\left( x \right)\cdot g\left( x \right)} \right)=f\left( x \right){g}’\left( x \right)+g\left( x \right){f}’\left( x \right)\) \(\displaystyle f\left( x \right)=5\left( {{{x}^{3}}-8x} \right)\left( {x-2} \right)\)

\(\displaystyle \begin{align}{f}’\left( x \right)&=5\left[ {\left( {{{x}^{3}}-8x} \right){{{\left( {x-2} \right)}}^{\prime }}+\left( {x-2} \right){{{\left( {{{x}^{3}}-8x} \right)}}^{{\prime }}}} \right]\\&=5\left[ {\left( {{{x}^{3}}-8x} \right)\left( {1-0} \right)+\left( {x-2} \right)\left( {3{{x}^{2}}-8} \right)} \right]\\&=5\left[ {{{x}^{3}}-8x+3{{x}^{3}}-8x-6{{x}^{2}}+16} \right]\\&=5\left[ {4{{x}^{3}}-6{{x}^{2}}-16x+16} \right]\\&=20{{x}^{3}}-30{{x}^{2}}-80x+80\end{align}\)

Quotient Rule \(\displaystyle \frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)=\frac{{g\left( x \right){f}’\left( x \right)-f\left( x \right){g}’\left( x \right)}}{{{{{\left( {g\left( x \right)} \right)}}^{2}}}}\)

\(\displaystyle g\left( x \right)\ne 0\)

\(\displaystyle f\left( x \right)=\frac{{{{x}^{2}}+1}}{{x+1}}\)

\(\displaystyle \begin{align}{f}’\left( x \right)&=\frac{{\left( {x+1} \right){{{\left( {{{x}^{2}}+1} \right)}}^{\prime }}-\left( {{{x}^{2}}+1} \right){{{\left( {x+1} \right)}}^{\prime }}}}{{{{{\left( {x+1} \right)}}^{2}}}}\\&=\frac{{\left( {x+1} \right)\left( {2x+0} \right)-\left( {{{x}^{2}}+1} \right)\left( {1+0} \right)}}{{{{{\left( {x+1} \right)}}^{2}}}}\\&=\frac{{2{{x}^{2}}+2x-{{x}^{2}}-1}}{{{{{\left( {x+1} \right)}}^{2}}}}=\frac{{{{x}^{2}}+2x-1}}{{{{{\left( {x+1} \right)}}^{2}}}}\end{align}\)

 

More Examples

Here are more examples. Do you see how you need to be really up on your algebra with calculus?

Function Derivative Notes
\(\displaystyle f\left( x \right)=\frac{1}{{\sqrt{{x+1}}}}\) \(\displaystyle \begin{align}f\left( x \right)&={{\left( {x+1} \right)}^{{-\frac{1}{2}}}}\\{f}’\left( x \right)&=-\frac{1}{2}{{\left( {x+1} \right)}^{{-\frac{3}{2}}}}=-\frac{1}{{2{{{\left( {\sqrt{{x+1}}} \right)}}^{3}}}}\end{align}\) Turn fraction with radical sign into expression with negative fractional exponent.

Note that you could use the quotient rule, but there is no need since we can simplify expression with an exponent, and just use the Power rule.

\(\displaystyle f\left( x \right)=4{{\pi }^{2}}\) \(\displaystyle \begin{array}{c}f\left( x \right)=\text{constant}\\{f}’\left( x \right)=0\end{array}\) You might think this might differentiate to \(8\pi \), but notice that there’s no \(x\) on the right-hand side, so the expression is actually a constant.

Use the Constant rule, and the derivative is 0.

\(\displaystyle f\left( x \right)=\frac{{3{{x}^{3}}-2{{x}^{2}}+1}}{{{{x}^{2}}}}\) \(\displaystyle \begin{align}f\left( x \right)&=3x-2+{{x}^{{-2}}}\\{f}’\left( x \right)&=3-2{{x}^{{-3}}}=3-\frac{2}{{{{x}^{3}}}}\end{align}\) It looks like we might have to use the quotient rule, but if we simplify the expression, we can just use the Power Rule and the Sum and Difference Rule.
\(\displaystyle f\left( x \right)=\left( {2x-4} \right)\left( {3x+1} \right)\) \(\displaystyle \begin{align}f\left( x \right)&=6{{x}^{2}}+2x-12x-4=6{{x}^{2}}-10x-4\\{f}’\left( x \right)&=12x-10\end{align}\) We don’t want to use the Product Rule since the coefficient of the \(x\)’s in the two factors aren’t 1 (we’ll learn later how to handle this with the Chain Rule). In this case, just multiply (FOIL) out the binomials and then use the Sum and Difference and Power Rules.
\(\displaystyle f\left( x \right)=\left( {\sqrt[4]{{{{x}^{5}}}}} \right)\left( {x+4} \right)\) We can multiply through first:

 

\(\displaystyle \begin{align}f\left( x \right)&={{x}^{{\frac{5}{4}}}}\left( {x+4} \right)={{x}^{{\frac{9}{4}}}}+4{{x}^{{\frac{5}{4}}}}\\{f}’\left( x \right)&=\frac{9}{4}{{x}^{{\frac{5}{4}}}}+5{{x}^{{\frac{1}{4}}}}=\frac{1}{4}\sqrt[4]{x}\left( {9x+20} \right)\end{align}\)

We get the same thing using the Product Rule:

\(\displaystyle \begin{align}{f}’\left( x \right)&={{x}^{{\frac{5}{4}}}}\left( 1 \right)+\left( {x+4} \right)\left( {\frac{5}{4}{{x}^{{\frac{1}{4}}}}} \right)\\&={{x}^{{\frac{5}{4}}}}+\frac{5}{4}{{x}^{{\frac{5}{4}}}}+5{{x}^{{\frac{1}{4}}}}\\&=\frac{9}{4}{{x}^{{\frac{5}{4}}}}+5{{x}^{{\frac{1}{4}}}}\end{align}\)

\(\displaystyle f\left( x \right)=\frac{{3{{x}^{3}}-2{{x}^{2}}+1}}{{{{x}^{2}}+2}}\) \(\displaystyle {f}’\left( x \right)\)

\(\displaystyle \begin{align}&=\frac{{\left( {{{x}^{2}}+2} \right)\left( {9{{x}^{2}}-4x} \right)-\left( {3{{x}^{3}}-2{{x}^{2}}+1} \right)\left( {2x} \right)}}{{{{{\left( {{{x}^{2}}+2} \right)}}^{2}}}}\\&=\frac{{9{{x}^{4}}-4{{x}^{3}}+18{{x}^{2}}-8x-\left( {6{{x}^{4}}-4{{x}^{3}}+2x} \right)}}{{{{{\left( {{{x}^{2}}+2} \right)}}^{2}}}}\\&=\frac{{3{{x}^{4}}+18{{x}^{2}}-10x}}{{{{{\left( {{{x}^{2}}+2} \right)}}^{2}}}}\end{align}\)

Since we can’t simplify this rational expression, we’ll use the Quotient Rule.
\(\displaystyle f\left( x \right)=\frac{{{{x}^{2}}-1}}{{x+1}}\) \(\require{cancel} \displaystyle \begin{align}f\left( x \right)&=\frac{{\left( {x-1} \right)\cancel{{\left( {x+1} \right)}}}}{{\cancel{{x+1}}}}=x-1\\{f}’\left( x \right)&=1\end{align}\) We could use the Quotient Rule, but we can also simplify first, and then take the derivative. Note that the function and derivative isn’t defined at \(x\ne -1\) (a removable discontinuity, or hole).

 

Here is another example of where we have use the Power Rule twice, since we’re multiplying three factors:

Function Derivative Using Power Rule Twice
\(\displaystyle \begin{array}{l}f\left( x \right)=x{{\left( {x+1} \right)}^{8}}{{\left( {x-2} \right)}^{3}}\\(f\left( x \right)=\left[ {x{{{\left( {x+1} \right)}}^{8}}} \right]{{\left( {x-2} \right)}^{3}})\end{array}\) \(\displaystyle \begin{align}{f}’\left( x \right)&=\left[ {x{{{\left( {x+1} \right)}}^{8}}} \right]\left[ {3{{{\left( {x-2} \right)}}^{2}}} \right]+{{\left( {x-2} \right)}^{3}}\cdot \left( {\text{derivative of }x{{{\left( {x+1} \right)}}^{8}}} \right)\\&=\left[ {x{{{\left( {x+1} \right)}}^{8}}} \right]\left[ {3{{{\left( {x-2} \right)}}^{2}}} \right]+{{\left( {x-2} \right)}^{3}}\left[ {x\cdot 8{{{\left( {x+1} \right)}}^{7}}+{{{\left( {x+1} \right)}}^{8}}\cdot 1} \right]\\&=3x{{\left( {x+1} \right)}^{8}}{{\left( {x-2} \right)}^{2}}+8x{{\left( {x+1} \right)}^{7}}{{\left( {x-2} \right)}^{3}}+{{\left( {x+1} \right)}^{8}}{{\left( {x-2} \right)}^{3}}\\&={{\left( {x+1} \right)}^{7}}{{\left( {x-2} \right)}^{2}}\left[ {3x\left( {x+1} \right)+8x\left( {x-2} \right)+\left( {x+1} \right)\left( {x-2} \right)} \right]\\&={{\left( {x+1} \right)}^{7}}{{\left( {x-2} \right)}^{2}}\left[ {3{{x}^{2}}+3x+8{{x}^{2}}-16x+{{x}^{2}}-x-2} \right]\\&={{\left( {x+1} \right)}^{7}}{{\left( {x-2} \right)}^{2}}\left[ {12{{x}^{2}}-14x-2} \right]\\&=2{{\left( {x+1} \right)}^{7}}{{\left( {x-2} \right)}^{2}}\left[ {6{{x}^{2}}-7x-1} \right]\end{align}\)

Here are some problems where you have use to the product and quotient rules to find derivatives at certain points using functions, or graphs of functions:

Product and Quotient Rule Derivative Problem Solution

Given that:

 

\(\begin{array}{c}g\left( 1 \right)=3\,\,\,\,\,\text{and}\,\,\,\,{g}’\left( 1 \right)=-1\\h\left( 1 \right)=-4\,\,\,\,\,\text{and}\,\,\,\,{h}’\left( 1 \right)=4\end{array}\)

 

 

Find \({f}’\left( 1 \right)\) when:

 

a) \(f\left( x \right)=3g\left( x \right)+2h\left( x \right)\)

b) \(f\left( x \right)=g\left( x \right)\cdot h\left( x \right)\)      

c) \(\displaystyle f\left( x \right)=\frac{{g\left( x \right)}}{{h\left( x \right)}}\)                 

a) \(\begin{array}{l}f\left( x \right)=3g\left( x \right)+2h\left( x \right)\\{f}’\left( x \right)=3{g}’\left( x \right)+2{h}’\left( x \right)\\{f}’\left( 1 \right)=3{g}’\left( 1 \right)+2{h}’\left( 1 \right)\\{f}’\left( 1 \right)=3\left( {-1} \right)+2\left( 4 \right)=5\end{array}\)               b) \(\displaystyle \begin{array}{l}f\left( x \right)=g\left( x \right)\cdot h\left( x \right)\\{f}’\left( x \right)=g\left( x \right)\cdot {h}’\left( x \right)+h\left( x \right)\cdot {g}’\left( x \right)\\{f}’\left( 1 \right)=g\left( 1 \right)\cdot {h}’\left( 1 \right)+h\left( 1 \right)\cdot {g}’\left( 1 \right)\\{f}’\left( 1 \right)=3\cdot 4+\left( {-4} \right)\left( {-1} \right)=16\end{array}\)

 

c) \(\displaystyle \begin{align}f\left( x \right)&=\frac{{g\left( x \right)}}{{h\left( x \right)}}\\{f}’\left( x \right)&=\frac{{h\left( x \right)\cdot {g}’\left( x \right)-g\left( x \right)\cdot {h}’\left( x \right)}}{{{{{\left( {h\left( x \right)} \right)}}^{2}}}}\\{f}’\left( 1 \right)&=\frac{{h\left( 1 \right)\cdot {g}’\left( 1 \right)-g\left( 1 \right)\cdot {h}’\left( 1 \right)}}{{{{{\left( {h\left( 1 \right)} \right)}}^{2}}}}=\frac{{\left( {-4} \right)\left( {-1} \right)-3\cdot 4}}{{{{{\left( {-4} \right)}}^{2}}}}=-\frac{1}{2}\end{align}\)

\(p\left( x \right)=f\left( x \right)\cdot g\left( x \right)\) and \(\displaystyle q\left( x \right)=\frac{{f\left( x \right)}}{{g\left( x \right)}}\)

Find \({p}’\left( 4 \right)\) and \({q}’\left( {-1} \right)\), given function \(f\) and \(g\) below:

Note the following (derivative is slope):

\(x\) \(f\left( x \right)\) \({f}’\left( x \right)\) \(g\left( x \right)\) \({g}’\left( x \right)\)
4 2 0 6 3
–1 2 2 3 –1
 

Find \({p}’\left( 4 \right)\):

 

\(\displaystyle \begin{align}p\left( x \right)&=f\left( x \right)\cdot g\left( x \right)\\{p}’\left( x \right)&=f\left( x \right)\cdot {g}’\left( x \right)+g\left( x \right)\cdot {f}’\left( x \right)\\{p}’\left( 4 \right)&=f\left( 4 \right)\cdot {g}’\left( 4 \right)+g\left( 4 \right)\cdot {f}’\left( 4 \right)\\{p}’\left( 4 \right)&=2\cdot 3+6\cdot 0=6\end{align}\)

 

 

Find \({q}’\left( {-1} \right)\):

 

\(\displaystyle \begin{align}q\left( x \right)&=\frac{{f\left( x \right)}}{{g\left( x \right)}}\\{q}’\left( x \right)&=\frac{{g\left( x \right)\cdot {f}’\left( x \right)-f\left( x \right)\cdot {g}’\left( x \right)}}{{{{{\left( {g\left( x \right)} \right)}}^{2}}}}\\{q}’\left( {-1} \right)&=\frac{{g\left( {-1} \right)\cdot {f}’\left( {-1} \right)-f\left( {-1} \right)\cdot {g}’\left( {-1} \right)}}{{{{{\left( {g\left( {-1} \right)} \right)}}^{2}}}}\\{q}’\left( {-1} \right)&=\frac{{3\cdot 2-2\left( {-1} \right)}}{{{{3}^{2}}}}=\frac{8}{9}\end{align}\)

Derivatives of Trig Functions

You basically just have to memorize the derivatives of the basic trig functions. Here they are:

Trigonometry Function Derivative Rule Hints
\(\displaystyle \frac{d}{{dx}}\left( {\sin \left( x \right)} \right)=\cos \left( x \right)\) Derivative of sine is cosine, derivative of cosine is –sine (basically just switch them).

 

Trig functions that start with “\(c\)” (like cosine) have negative derivatives.

\(\displaystyle \frac{d}{{dx}}\left( {\cos \left( x \right)} \right)=-\sin \left( x \right)\)
\(\displaystyle \frac{d}{{dx}}\left( {\tan \left( x \right)} \right)={{\sec }^{2}}\left( x \right)\) Each of these derivative statements has either one tangent and two secants, or one cotangents and two cosecants.

 

For example, the derivative of tangent is \(\displaystyle \text{secan}{{\text{t}}^{2}}\) (one tangent and two secants), and the derivative of secant is \(\text{secant}\times \text{tangent}\) (one tangent and two secants).

 

Similarly, the derivative of cotangent is \(-\text{cosecan}{{\text{t}}^{2}}\) (one cotangent and two cosecants), and the derivative of cosecant is \(-\text{cosecant}\times \text{cotangent}\) (one cotangent and two cosecants).

 

Again, note that trig functions that start with “\(c\)”  (cotangent and cosecant) have negative derivatives.

 

 

\(\displaystyle \frac{d}{{dx}}\left( {\cot \left( x \right)} \right)=-{{\csc }^{2}}\left( x \right)\)
\(\displaystyle \frac{d}{{dx}}\left( {\sec \left( x \right)} \right)=\sec \left( x \right)\tan \left( x \right)\)
\(\displaystyle \frac{d}{{dx}}\left( {\csc \left( x \right)} \right)=-\csc \left( x \right)\cot \left( x \right)\)

Here are a few examples; note that the sum and difference rules (and all the other rules) apply for trig derivatives, too:

Trig Function Derivative Notes
\(\displaystyle f\left( x \right)=3\sin x-2\cos x\) \(\displaystyle \begin{align}{f}’\left( x \right)&=3\cdot \cos x-2\cdot \left( {-\sin x} \right)\\&=3\cos x+2\sin x\end{align}\) Per the Sum and Difference Rule, take the derivative of each trig function separately and subtract, just like the original function. It’s as easy as that!
\(\displaystyle f\left( x \right)=3x\cos x-{{x}^{2}}\sec x\) \(\displaystyle \begin{align}{f}’\left( x \right)&=\left( {3x} \right)\left( {-\sin x} \right)+\left( {\cos x} \right)\left( 3 \right)\\&\,\,\,\,\,\,\,\,\,-\left[ {{{x}^{2}}\left( {\sec x\tan x} \right)+\sec x\left( {2x} \right)} \right]\\&=-3x\sin x+3\cos x-{{x}^{2}}\sec x\tan x-2x\sec x\end{align}\) This one’s a little trickier since we need to use the Product Rule (twice) in addition to the Sum and Difference Rule. Note in this example, we took the first part of the function as “\(3x\)” instead of leaving the 3 on the outside and distributing later. Both methods work.

 

Higher Order Derivatives

We can actually take the derivative of a function more than once; we’ll see this here in the Equation of the Tangent Line, Tangent Line Approximation, and Rates of Change Section. We’ll see that the acceleration function is the derivative of the velocity function, which is the derivative of the position function. (In this case, we would take the derivative twice of the position function to get the acceleration function.)

The second derivative (and third derivative, and so on) is what we call a higher order derivative, and the notation looks like the following:

First Derivative \(\displaystyle {y}’\text{,}\,\,\,\,\,\,\,\,\,\,\,\,{f}’\left( x \right),\,\,\,\,\,\,\,\,\,\,{F}’\left( x \right),\,\,\,\,\,\,\,\,\,\,\,\frac{{dy}}{{dx}},\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {f\left( x \right)} \right],\,\,\,\,\,\,\,\,\,\,{{D}_{x}}\left[ y \right]\)
Second Derivative \(\displaystyle {y}^{\prime \prime}\text{,}\,\,\,\,\,\,\,\,\,\,\,\,{f}^{\prime \prime}\left( x \right),\,\,\,\,\,\,\,\,\,\,{F}^{\prime \prime}\left( x \right),\,\,\,\,\,\,\,\,\,\,\,\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}},\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{{d}^{2}}}}{{d{{x}^{2}}}}\left[ {f\left( x \right)} \right],\,\,\,\,\,\,\,\,\,\,{{D}_{x}}^{2}\left[ y \right]\)
Third Derivative \(\displaystyle {y}^{\prime \prime \prime}\text{,}\,\,\,\,\,\,\,\,\,\,\,\,{f}^{\prime \prime \prime}\left( x \right),\,\,\,\,\,\,\,\,\,\,{F}^{\prime \prime \prime}\left( x \right),\,\,\,\,\,\,\,\,\,\,\,\frac{{{{d}^{3}}y}}{{d{{x}^{3}}}},\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{{d}^{3}}}}{{d{{x}^{3}}}}\left[ {f\left( x \right)} \right],\,\,\,\,\,\,\,\,\,\,{{D}_{x}}^{3}\left[ y \right]\)
Fourth Derivative \(\displaystyle {{y}^{{\left( 4 \right)}}}\text{,}\,\,\,\,\,\,\,\,\,\,\,\,{{f}^{{\left( 4 \right)}}}\left( x \right),\,\,\,\,\,\,\,\,\,\,{{F}^{{\left( 4 \right)}}}\left( x \right),\,\,\,\,\,\,\,\,\,\,\,\frac{{{{d}^{4}}y}}{{d{{x}^{4}}}},\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{{d}^{4}}}}{{d{{x}^{4}}}}\left[ {f\left( x \right)} \right],\,\,\,\,\,\,\,\,\,\,{{D}_{x}}^{4}\left[ y \right]\)
\(n\)th Derivative \(\displaystyle {{y}^{{\left( n \right)}}}\text{,}\,\,\,\,\,\,\,\,\,\,\,\,{{f}^{{\left( n \right)}}}\left( x \right),\,\,\,\,\,\,\,\,\,\,{{F}^{{\left( n \right)}}}\left( x \right),\,\,\,\,\,\,\,\,\,\,\,\frac{{{{d}^{n}}y}}{{d{{x}^{n}}}},\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{{d}^{n}}}}{{d{{x}^{n}}}}\left[ {f\left( x \right)} \right],\,\,\,\,\,\,\,\,\,\,{{D}_{x}}^{n}\left[ y \right]\)

(Don’t let all this scare you; you’ll usually just be using one or two types of notation, and it will be pretty obvious).

And with polynomials, if we keep taking derivatives, we’ll eventually end up with 0; for example, for \(y=5{{x}^{4}}-3{{x}^{2}}+2x+3\),  we have \({y}’=20{{x}^{3}}-6x+2,\,\,\,\,\,{y}^{\prime \prime}=60{{x}^{2}}-6;\,\,\,\,\,{y}^{\prime \prime \prime}=120x;\,\,\,\,\,\,{{y}^{{\left( 4 \right)}}}=120;\,\,\,\,\,\,{{y}^{{\left( 5 \right)}}}=0\).

Here are more problems; note that in the first case, we have what we call “indestructible derivatives” with the sin and cos, since we can keep taking the derivative forever, and the functions never go away (you won’t end up with 0, as in the previous example with a polynomial).

Trig Function 1st and 2nd Derivatives Notes
\(\displaystyle f\left( x \right)=3\sin x-2\cos x\) \(\displaystyle \begin{align}{f}’\left( x \right)&=3\cos x-2\left( {-\sin x} \right)\\&=3\cos x+2\sin x\end{align}\)

 

\(\displaystyle \begin{align}{{f}^{\prime \prime}(x)} &=3\left( {-\sin x} \right)+2\cos x\\&=-3\sin x+2\cos x\end{align}\)

Use the same rules over and over again. Do you see how with these functions (called “indestructible derivatives“) never “go away” (end up with being 0) after taking the derivative over and over again?
\(\displaystyle f\left( x \right)=\frac{{{{x}^{2}}+2x}}{{x-3}}\)

\(\displaystyle \begin{align}{f}’\left( x \right)&=\frac{{\left( {x-3} \right)\left( {2x+2} \right)-\left( {{{x}^{2}}+2x} \right)\left( 1 \right)}}{{{{{\left( {x-3} \right)}}^{2}}}}\\&=\frac{{2{{x}^{2}}-4x-6-{{x}^{2}}-2x}}{{{{{\left( {x-3} \right)}}^{2}}}}=\frac{{{{x}^{2}}-6x-6}}{{{{{\left( {x-3} \right)}}^{2}}}}\end{align}\)

 

\(\require {cancel} \displaystyle \begin{align}{{f}^{\prime \prime}(x)}&=\frac{{{{{\left( {x-3} \right)}}^{2}}\left( {2x-6} \right)-\left( {{{x}^{2}}-6x-6} \right)\cdot 2\left( {x-3} \right)}}{{{{{\left( {x-3} \right)}}^{4}}}}\\&=\frac{{2{{{\left( {x-3} \right)}}^{3}}-2\left( {x-3} \right)\left( {{{x}^{2}}-6x-6} \right)}}{{{{{\left( {x-3} \right)}}^{4}}}}\\&=\frac{{2\cancel{{\left( {x-3} \right)}}\left[ {{{{\left( {x-3} \right)}}^{2}}-\left( {{{x}^{2}}-6x-6} \right)} \right]}}{{{{{\left( {x-3} \right)}}^{{\cancel{4}}}}^{3}}}\\&=\frac{{2\left( {\cancel{{{{x}^{2}}}}-\cancel{{6x}}+9-\cancel{{{{x}^{2}}}}+\cancel{{6x}}+6} \right)}}{{{{{\left( {x-3} \right)}}^{3}}}}=\frac{{30}}{{{{{\left( {x-3} \right)}}^{3}}}}\end{align}\)

 

We have to use the Quotient Rule twice; it’s a lot of messy algebra, but not too bad.
\(\displaystyle f\left( x \right)=-\sec x\) \(\displaystyle {f}’\left( x \right)=-\sec x\tan x\)

 

\(\displaystyle \begin{align}{{f}^{\prime \prime}(x)}&=-\left[ {\sec x\cdot {{{\sec }}^{2}}x+\tan x\cdot \left( {\sec x\cdot \tan x} \right)} \right]\\&=-{{\sec }^{3}}x-\sec x\cdot {{\tan }^{2}}x\\&=-\sec x\left( {{{{\sec }}^{2}}x+{{{\tan }}^{2}}x} \right)\end{align}\)

In this one, we have to use the Product Rule for the second derivative, since the first derivative ended up with the product of two trig functions.

 

Note that we factored the sec out for the last step; this would be optional.

 

Understand these problems, and practice, practice, practice!


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On to Equation of the Tangent Line, Tangent Line Approximation, and Rates of Change – you’re ready!

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