| Constant Rule | Examples of Constant, Power, Product and Quotient Rules |
| Power Rule | Derivatives of Trigonometric Functions |
| Product Rule | Higher Order Derivatives |
| Quotient Rule | More Practice |
| List of Rules |
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. Note that there are examples for all these rules here.
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.
Remember that with 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.
(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 in the Exponential and Logarithmic Differentiation section.)
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, for $ y=3$ , the derivative of that function is 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; otherwise it gets more complicated (although we can always turn something like $ {{\left( {2x} \right)}^{5}}$ into $ 32{{x}^{5}}$).
As an example of the Power Rule, for $ y=4{{x}^{3}}$, 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, the derivative of the volume of a sphere with respect to its radius, $ \displaystyle V=\frac{4}{3}\pi {{r}^{3}}$ is 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! 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 a certain value such as $ 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, since the derivative of the product of two expressions with variables in them is NOT the product of their derivatives! 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, turn exponents on the bottom into negative exponents; for example, $ \displaystyle \frac{5}{{{{x}^{2}}}}=5{{x}^{{-2}}}$. Bottom line: always try to simplify the function first.
Here’s how I like to remember it: the derivative is first times the derivative of the second PLUS second times derivative of the first. Yes, that is a plus (addition) in the middle. 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. Here is what it looks like in Theorem form (you may learn it a little differently, like in a different order, but it will come out the same):
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:
$ \displaystyle \begin{align}\frac{d}{{dx}}\left[ {5x{{{\left( {x+1} \right)}}^{3}}} \right]&=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}\frac{d}{{dx}}\left[ {5x{{{\left( {x+1} \right)}}^{3}}} \right]&=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 (like we learned here in the Advanced Factoring section), 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 (constants) in the denominator – only if there are variables 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}}$). And you can always turn a denominator into a numerator using negative exponents, but typically the quotient rule is easier. For example, $ \displaystyle \frac{x}{{{{x}^{3}}+3}}=x{{\left( {{{x}^{3}}+3} \right)}^{{-1}}}$:
Product Rule: $ \displaystyle \begin{align}\frac{{d\left( {\frac{x}{{{{x}^{3}}+3}}} \right)}}{{dx}}&=\frac{{d\left[ {x{{{\left( {{{x}^{3}}+3} \right)}}^{{-1}}}} \right]}}{{dx}}=x\cdot -1{{\left( {{{x}^{3}}+3} \right)}^{{-2}}}\cdot 3{{x}^{2}}+{{\left( {{{x}^{3}}+3} \right)}^{{-1}}}\cdot 1\\&={{\left( {{{x}^{3}}+3} \right)}^{{-2}}}\left[ {-3{{x}^{3}}+\left( {{{x}^{3}}+3} \right)} \right]=\frac{{3-2{{x}^{3}}}}{{{{{\left( {{{x}^{3}}+3} \right)}}^{2}}}}\end{align}$
Quotient Rule: $ \displaystyle \frac{{d\left( {\frac{x}{{{{x}^{3}}+3}}} \right)}}{{dx}}=\frac{{\left( {{{x}^{3}}+3} \right)\cdot 1-x\cdot 3{{x}^{2}}}}{{{{{\left( {{{x}^{3}}+3} \right)}}^{2}}}}=\frac{{3-2{{x}^{2}}}}{{{{{\left( {{{x}^{3}}+3} \right)}}^{2}}}}$
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 Lo De Hi way to remember (Lo dHi minus Hi dLo, all over Lo Lo, or the square of what’s beLo):
$ \displaystyle \frac{d}{{dx}}\left( {\frac{{Hi}}{{Lo}}} \right)=\frac{{Lo\cdot dHi-Hi\cdot dLo}}{{Lo\cdot Lo}}$
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).
| Differentiation 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}$ |
| Sum and Difference Rule | $ \displaystyle \frac{d}{{dx}}\left[ {f\left( x \right)\pm g\left( x \right)} \right]=\frac{d}{{dx}}f\left( x \right)\pm \frac{d}{{dx}}g\left( x \right)$ | $ f\left( x \right)={{x}^{2}}-x+3$ $ \displaystyle {f}’\left( x \right)=2x-1$ |
More Examples
Here are more examples. Here are more examples. Do you see how important algebra is 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 the Quotient Rule could have been used, but there is no need because of the simplification. |
| $ \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 use the Sum and Difference, Constant, and Power Rules. |
| $ \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 can’t use the Product Rule as is 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 for composition of functions). In this case, multiply (FOIL) the binomials and then use the Sum and Difference, Constant, and Power Rules. |
| $ \displaystyle f\left( x \right)=\left( {\sqrt[4]{{{{x}^{5}}}}} \right)\left( {x+4} \right)$ | Its easiest to 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, 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}$ | |||||||||||||||
| Given $ 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):
| 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 Trigonometric Functions
You basically just have to memorize the derivatives of the Basic Trigonometric Functions. Here they are:
| Trigonometry Function Derivative Rule | Hints |
| $ \displaystyle \frac{d}{{dx}}\left( {\sin x} \right)=\cos x$ $ \displaystyle \frac{d}{{dx}}\left( {\cos x} \right)=-\sin x$ $ \displaystyle \frac{d}{{dx}}\left( {\tan x} \right)={{\sec }^{2}}x$ $ \displaystyle \frac{d}{{dx}}\left( {\cot x} \right)=-{{\csc }^{2}}x$ $ \displaystyle \frac{d}{{dx}}\left( {\sec x} \right)=\sec x\cdot \tan x$ $ \displaystyle \frac{d}{{dx}}\left( {\csc x} \right)=-\csc x\cdot \cot x$ | Derivative of sine is cosine, derivative of cosine is –sine (basically just switch them).
For the trig functions that aren’t sine or cosine, each 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).
Note that trig functions that start with “$ c$” (cosine, cotangent and cosecant) have negative derivatives. |
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. |
| $ \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. In the first term, we took the first part of the function as “$ 3x$” instead of leaving the $ 3$ on the outside and distributing later. Both methods work. |
| $ \displaystyle f\left( x \right)={{x}^{3}}\cot x+\frac{1}{2}\csc x$ | $ \displaystyle \begin{align}{f}’\left( x \right)&=\left( {{{x}^{3}}} \right)\left( {-{{{\csc }}^{2}}x} \right)+\left( {\cot x} \right)\left( {3{{x}^{2}}} \right)\\&\,\,\,\,\,\,\,\,\,+\frac{1}{2}\left( {-\csc x\cot x} \right)\\&=-{{x}^{3}}{{\csc }^{2}}x+3{{x}^{2}}\cot x-\frac{1}{2}\csc x\cot x\end{align}$ | Note that the derivatives of the trig functions that begin with “$ c$” are negative. Also remember that there are always two csc’s and one cot in the csc/cot derivatives (derivative of cot is $ -{{\csc }^{2}}$ and derivative of csc is $ -\csc \cot $). |
| $ \displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\cos \left( {x+h} \right)-\cos x}}{h}$ | $ -\sin x$ | Since $ \displaystyle {f}’\left( x \right)=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}$ (from the Definition of the Derivative section), this is just the derivative of $ \cos x$, which is $ -\sin x$. |
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}’,\,\,{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$, $ {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}$
| 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, use the Product Rule for the second derivative, since the first derivative ended up with the product of two trig functions. |
Understand these problems, and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.
If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets.
You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!
On to Equation of the Tangent Line, Tangent Line Approximation, and Rates of Change – you’re ready!