# Math Help - Derivative Tutorial

1. ## Derivative Tutorial...Feel free to thank on any post!!

I used this in a recent post and maybe thought it would be beneficial to some people to have it posted as a tutorial...hope it helps!

Basic rules

where c is a constant

...which implies that if the base is e then the derivative is

now we get to some rules

Product rule
where u and v are functions of x

Quotient rule

Chain rule

Now that I have discussed both I will talk about the fact that sometimes easier than the quotient rule is applying the chain rule to quotients by using the fact taht ...therefore by the chain rule

Another very useful technique is logarithmic differentiatiow which takes full advantage of logarithims ability to simplify complex functions...this one comes in two flavors...simplification of large quotients...or differentiating when there is an x in the exponent and in the base

I will give you a general case for the second one ..now applying our logarithims(prefably always ln(x) since it is the nicest to work with) we get ...now using the amazing simplification techniques of logarithim this turns into ...now differentiating we get ...the right side was gotten using the fact that and the left side was gotten using the product rule...and we multiply both sides by y to get the final answer remember that

for the first use I willshow you an example ..now using that tecnhique we have ...simplifying we get ...differentiating we get ...now just multiply both sides by y(the original function) and you have your answer...I will post some practice questions you can work and we can critique you

2. These are the questions mentioned in my post...I will gladly help anyone that wishes to get some practice by solving them and submitting their answers...this is open to anyone who wishes to participate!

Mathstud

These are NOT in order in which I covered them to make them more challenging

Find the derivatives of the following equations
1.

2.

3.

4.

5.

6.

7.

8.

You can do these if you wish and we will critique you if you post them...hope I have been some help!

3. And for those who are a bit more adventurous....Try this

Find y'

$y=2^{\ln(\frac{1}{x^3})}+x^{\ln\bigg(\frac{arcsin( x^2)}{cos(x)}\bigg)}$ I have the answer is someone wishes to check theirs!

4. In conjunction with this I will just put the rest of what I did in the other post here so it can be used as a "quick study guide" to remember the major points in Calc AB

Part 1 useful integral tricks

remember taht
O and three VERY helpful identities ,

,

another thing is you always want to get integrals into the form

Integral's using parts

Some integrals that dont seem intuitive are found using integration by parts and calling $dv=1\cdot{dx}$...two examples

$\int{ln(ax)dx}$

calling $u=ln(ax)$ $dv=1\cdot{dx}$

we get $du=\frac{a}{ax}=\frac{1}{x}dx$ and $v=x$

so no applying parts $\int{u\cdot{dv}}=u\cdot{dv}-\int{v\cdot{du}}$ to get $\int{\ln(ax)dx}=x\cdot\ln(ax)-\int\frac{1}{x}\cdot{x}dx=x\cdot{ln(ax)}-x$

and the other is arctan(x).

Calling $u=arctan(x)$ and $dv=1\cdot{dx}$

you get $du=\frac{1}{1+x^2}dx$ and $v=x$

so applying the same formula we get $\int{arctan(x)}dx=x\cdot{arctan(x)}-\int\frac{x}{1+x^2}dx=x\cdot{arctan(x)}-\frac{1}{2}ln|x^2+1|+C$

Two more useful suggestions are when the numerator is one degree higher or equal to the denominator you should polynomial divide or the equivalent o get an easier integral....example $\int\frac{x^2}{x^2+1}dx$...using polynomial division or this equivalent $\int\frac{x^2}{x^2+1}dx=\int\frac{x^2+1-1}{x^2+1}dx=\int{1-\frac{1}{x^2+1}dx=x-arctan(x)+C}$

The last thing I want to suggest is LEARN PARTIAL FRACTIONS DECOMPOSTION...you will not regret it...maybe I will post some other later

5. Hello,

LEARN PARTIAL FRACTIONS DECOMPOSTION
And then (or learn before) division of polynomials.

For partial fractions decomposition, the essential thing to know is that if we work in $\mathbb{R}$, any polynomial is a product of polynomials of degree 1 and 2. Then, if you decompose in partial fractions, you'll have either the derivative of a logarithm, either the derivative of arctan.

Dunno if it'll be useful.

6. 2. this is about the commonly asked question of tangent or normal lines to a point...also this deals with the blissful theorem of L'hopitals

Sorry I am posting up a storm...but another useful bit of info would be

Equation of a tangent line to the point

Equation of normal line at point

and I know everyone hates it but L'hopital's rule

which states that if the limit gives one of the 8 or so indeterminate forms(most commonly or ...then

The other useful indeterminate forms are $1^{\infty},0^{0},0\cdot{\infty},0^{\infty}$

a good example of a case where L'hopital's rule is useful but admittedly not neccasary is $\lim_{\Delta{x} \to 0}\frac{1-cos\bigg(\Delta{x}\bigg)}{\Delta{x}}$...now applying L'hopital's since this is an indeterminate form we get $\lim_{\Delta{x} \to 0}\frac{1-cos\bigg(\Delta{x}\bigg)}{\Delta{x}}=\lim_{\Delta{ x} \to 0}\frac{-sin\bigg(\Delta{x}\bigg)}{1}=0$ this occurs in the formal defnition of the derivative of sin(x) usign the difference quotient

another useful method for differentiating comes from once again using logarithims...example $y=\lim_{x \to 0}(1+x)^{\frac{1}{x}}$

taking the ln of both sides we get $ln(y)=lim_{x \to 0}\frac{1}{x}\cdot\ln(1+x)$...now applying L'hopitals' rule to the right side since it is an indeterminate form of $\frac{0}{0}$ we rewrite it as $\lim_{x \to 0}\frac{ln(1+x)}{x}=\lim_{x \to 0}\frac{\frac{1}{1+x}}{1}=\lim_{x \to 0}\frac{1}{1+x}=1$ but remembering that this is actually $\ln(y)=1\Rightarrow{y=e}$....therefore $\lim_{x \to 0}(1+x)^{\frac{1}{x}}=e$ which is the definition of $e\approx{2.71828}$

7. 3. Some theorems that you need to apply often
Commonly thrown at you are the following

Rolle's Theorem "If f is continous on [a,b] and differentiable on (a,b), and f(a)=f(b) then there exists one value c between a and b such that "...which basically says that if on an interval a function meets the requirements of the theorem there is a max or a min

Mean-Value theorem "If f is continuous on [a,b] and differentiable on (a,b). then there exists one value c between a and b such that which basically says that there is one point on an interval where the actual slope is equal to the average slope(i.e. )

Extreme Value theorem
"If f is continous on [a,b] then there exists an absolute maximum point and an absolute minimum point"

that one is self-explanitory

These make sense to anyone I think

8. Dunno what to say...

colby made calculus notes on MHF and released that here : The Calculus Guru

9. 4. Methods for finding Relative extrema/inflection points

First derivative test:
Find the points that make the first derivative 0, the first derviative undefined, and the points that make the original function undefined...set up test intervals based on theses numbers....test an element of each interval in the first derivative...where the sign's of the evaluated points change from + to - it is a max...and where they change from - to+ is a min

Second derivative test:
Take ONLY the points that make the first derivative 0...plug those points into the second derivative if $f''(c)>0$ c is a min....if [tex]f''(c)<0[\math] c is a max...if then the test is inconclusive...c is now a candidate for an inflection point

Test for inflection points:
Find the points where the second derivative is undefined, the points where the second dervative is equal to 0, and the points that make the original function undefined...Set up test intervals based on theses numbers...test an element of each interval in the second derviative...where there is a sign change there is an inflection point

NOTE: One thing many students forget to discern between is relative extrema and absolute extrema....Relative exterma occur at four places...where the first derivative is 0, the first derivative is undefined, the original function is undefined, and the enpoints of the domain(i.e. the endpoints they ask you to ind the interval on...the 0 and 10 in $0\leq{x}\leq{10}$)...but Absolute extrema occur in only three places...where the derivative is 0, where the derivative is undefined, and the endpoints of the domain...if you tried the points that make the original function undefined...you would be including points that don't actually exist...and to find which is the absolute max/min test all the relative maxs/mins in the original function whichever yields the smallest value is the absolute min and the largest value yielded is the relative max

10. 5. Differentiating inverse functions

One topic not covered in the aforementioned site is the derivatives of inverse functions...ok this is the best way I have discovered...I think it is right...it has always worked out before...I made it up so if its wrong I am sorry......therefore is the inverse....now taking the derivative we get ..but we want ...so....we reciprocate!....to get

another topic that should probably be covered is how to find the inverse......here is how you would do it $y=ln(x^2-4x+4)$...to find the inverse first switch the x's and y's to get $x=ln(y^2-4y+4y)$..now solve for x again to get the inverse $e^x=y^2-4y+4y=(y-2)^2$...therefore $e^{\frac{x}{2}}+2=y$...now this is a good test....if g(x) is an inverse of f(x) then it doesnt necasarrily follow that $f(g(x))=x$ and $g(f(x))=$...example $x^2$ and $\sqrt{x}$...but if it does you know it is an inverse...make sure that your inverse is one-to-one...if it is not you need to make a domain restrictions...for example the inverse of $\cos(x)$ is $arcos(x)$ which is only defined on $0\leq{x}\leq{\pi}$...this is because arcos(x) is only one-to-one or bijective on that interval...same goes for artan(x) and arcsin(x)...whose restricted domains are $\frac{-\pi}{2} and $\frac{-\pi}{2}\leq{x}\leq\frac{\pi}{2}$ respectively

11. Now I will show another method for integration that takes some experience to recognize....say you had this integral $\int\frac{1}{e^{x}+1}dx$...you look at that and go..."Hmm...arctan...no....ln(e^x+1)..hmm...not that either" here is what you do...since you can technically add 0 to anything and remain equality you can rewrite this integral as $\int\frac{1+e^{x}-e^{x}}{1+e^{x}}dx\Rightarrow\int\frac{e^{x}+1}{e^{ x}+1}dx-\int\frac{e^{x}}{e^{x}+1}dx$..which is obviously $x-\ln(e^{x}+1)+C$....another good example of this is done by Krizalid in this post http://www.mathhelpforum.com/math-he...-integral.html

12. Why do you call this a derivative tutorial if you deal with integrals also?

13. Originally Posted by Jhevon
Why do you call this a derivative tutorial if you deal with integrals also?
By the way, I thought to make a thread with integration techniques.

14. But I don't even start yet, I'll post it in Calculus forum when having some of time.

15. Originally Posted by Mathstud28
Can't wait!
I second that!

and please, start simple with us lesser people, Krizalid

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