<p>Thanks for the awesome limit study guide. We finished a week or so ago but I know I’ll use it come midterms. Glad to here everyone’s doing well.</p>
<p>Senior, AB</p>
<p>We’ve just started “Determining if a Function is Differentiable”.</p>
<p>I will definitely be using that limit study guide for the first nine weeks exam. Thanks.</p>
<p>no problem, I will make lists based the units of my teacher’s curriculum map.
Just as a sign for the future, here are all the lists I will be making
2) Introduction to Differentiation
3) The Chain Rule/Higher order of Derviatives
4) Mean Value Theorem and applications
5) Related Rates/L’Hospital’s Rule
6) Intro to Integration + integration with natural logs
7) More Integration + U substituion
8) Applications of calculus with integregation by parts
9) Transcendatal Functions
10) Volume (Shell, Disc, and Washer) (I don’t know what that means)</p>
<p>Im taking AP Calc AB!!! W0000, except first chapter test i made stupid mistake on limits and trapezoid area rule…</p>
<p>like it was limit of cos(1/x) as x approaches infinity… i put Does not exist LOL!</p>
<p>But yeah, we’re learning definition of derivative and anti-derivatives right now</p>
<p>CSMath, I have a test on limits tomorrow and I just looked at your guide which was actually really helpful. Definitely cleared up a lot of things I didn’t get</p>
<p>No problem, I am glad I helped jspede12. I am almost done with my dervatives one also.</p>
<p>I hope you will put it up soon. I got a huge derivative test next week. I’m definitely going to need any help I can get.</p>
<p>The thing my teacher keeps telling us with derivatives is to write out all the rules we need to use in the proplem before solving it, so then it just becomes a complex algerbra proplem. In my class most of the issuess have been people making mistakes simplifing them and not with the actual calculus as my teacher puts it. Just out of curiousity what book is everyone using? My school uses the Ron Larson 8 edition book.</p>
<p>Hi Guys! Here is the study guide for Intro to Differentiation:
Differentiation - process for finding a derivative (slope)
Notation for derivatives (be familiar with all notations)
f’(x) (read as f prime of x) (d/dx)[f(x)) (derivate f(x) with respect to x)
y’ (y prime) dy/dx (change in y over change in x)
Dx[y]
All of these notations mean differentiation
Slope formula = (y2-y1/ x2-x1)
This basic formula is basis for the formula
f’(x) = f(x+h) -f(x) / (h) (h means change in x)
lim h–>0<br>
Those 2 equations are the same thing
The Derivative of a linear function
This should be easy, you can find the derivative of a linear function easily because the function is in y=mx+b (m is slope)
Just to prove it to you, I will use the limit process to find the derivative
f(x) = 2x-3 (the slope or derivative should be 2)
f’(x)= f(x+h) - f(x)/h
lim h–> 0<br>
(now when you put the equation in for the function (x+h) becomes your new “x” and for f(x) you just substitute the function
f’(x)= (2(x+h)-3 - [2x-3])/h
lim h–> 0 (remember you are subtracted the entire function 2x-3, not just 2x)
f’(x) = (2x+2h-3 -2x+3)/h
lim h–> 0<br>
f’(x) = 2h/h =2 (2 is your derivative)<br>
lim h–> 0<br>
Derivatives of a nonlinear function
f(x)= 3x^2-4
f’(x)= (f(x+h) - f(x))/h
lim h–>0
(3(x+h)^2-4 - [3x^2-4])/h (substitute the function in)
lim h–>0
(3(x^2+2xh+h^2)-4 -3x^2+4)/h (expand (x+h)^2)
lim h–>0
(3x^2+6xh+3h^2-4-3x^2+4)/h (multiply 2 to the the expansion of (x+h)^2
lim h–>0
(6xh+3h^2)/h (get rid of the terms that are opposites
lim h–>0
h(6x+3h)/h (factor out an h)
lim h–>0
6x+3h (now use direct substitution)
lim h–>0
6x = your derivative for 3x^2-4</p>
<p>(Note)
f’(x) = lim (f(x+h) - f(x))/h (the limit process for a derivative)
h–>0<br>
This process to find a derivative always works provided that the limit exists. </p>
<p>Differentiating a cubic function with the limit process
x^3+2x
lim (f(x+h) - f(x))/h (write difference quotient)
lim h–>0
lim ((x+h)^3+2(x+h) - [x^3+2x])/h (we substitute numbers in)
lim h–>0
lim (x^3+3x^2h+3xh^2+h^3 +2x+2h- x^3-2x)/h
h–> 0 (expand, note you can either multiply the trinomial (x+3)(x+3)(x+3) three times(this is annoying especially with higher powers, you can use Pascal’s triangle, or you can use binomial theorem. (I won’t explain either of those things because those are precalc topics and I am discussing calculus. Also, in a few sections down, we won’t be using this process anyway)
lim (3x^2h+3xh^2+h^3+2h)/h (get rid of opposite terms)
h–>0
lim h(3x^2+3xh+h^2+2)/h (factor out h)
h–>0
lim 3x^2+3x(0)+(0)^2+2 (direct substitution)
h–>0
f’(x) =3x^2 +2</p>
<p>Deviating using “conjugate base fancy 1 way”
f(x) = x^(1/2)
Note (you should know this but x^(1/2) is the square root of x)
lim f(x+h) -f(x) / (h)
h–>0<br>
lim ((x+h) ^(1/2) - [x^(1/2)])/h (substitute in function)
h–>0
lim (((x+h) ^(1/2) - [x^(1/2)])/h) ((x+h)^(1/2) +x^(1/2)])/((x+h)^(1/2) +x^(1/2)]) (if you remember your algebra 2/trig then you would multiply this function by a its conjugate base and it has to technically be a 1 ((x+h)^(1/2) +x^(1/2)])/((x+h)^(1/2) +x^(1/2)]) is equal to 1 and it is the conjugate base to (((x+h) ^(1/2) - [x^(1/2)])/h) (a conjugate base is like a+b to a-b)
h–>0
Lim x+h-x/(h((x+h)^(1/2)+x^(1/2)) (simplify)
h–>0
Lim h/ (h((x+h)^(1/2)+x^(1/2))(take opposites (h and -h)
h–>0
1/((x+h)^(1/2)+x^(1/2)) (factor out an h)
h–>0
1/(x) ^ (1/2) + x^ (1/2) (simplify)
f’(x)= 1/2x^(1/2) (combine like terms)</p>
<p>Derivate using “common denominator fancy 1 way”
y = 3/x
lim f(x+h) - f(x)/h (difference quotient)
h–>0
lim (3/(x+h)) - [3/x]/h (substitute)
h –>0
lim (3/(x+h) (x/x) - <a href=“x+h”>3/x</a>/(h) )/ h (multiply by fancy 1s to get a common denominator
h –>0
lim (3x/x(x+h) - (3(x+h)/ x(x+h))/h (get new fraction)
h –>0
lim (3x/x(x+h) - (3x+3h)/ x(x+h))/h (multiply)
h –>0
lim (3x - [3x+3h]/ x(x+h) )/h (combined fraction)
h –>0
lim (-3h/x(x+h))/h (get rid of opposites)
h–>0
lim -3/ x(x+h) (factor out an h)
h–>0
lim -3/ (x)(x+0) (direct sub)
h –>0
lim -3/ x^2
h –>0
f’(x) = -3/x^2 (here is the derivative)
WAM Theorem- If a function is differentiable at x=c, then it is continuous at x=c. So, differentiability implies continuity
SHORT CUTS!!!
Constant rule - the derivative of a constant is 0
dy/dx[3] =0, why? graph y=3 for example it has a 0 slope, so the derivative of 3 is 0
Power Rule:
Proof: d/dx x^n = nx^(n-1)<br>
Note this proof involves combinaticrics so if you don’t remember that from precalc, then you won’t get the proof. (n 1) “means n choose 1”
lim (x+h) ^n - x^n
h–>0
lim (x^n +(n 1)x^(n-1)h +(n 2) x^(n-2)h^2… (n n) x^0 h^n - x^n)/h
h–>0
divide everthing by h
you will be left with x^(n-1) because all the other terms will be 0 after direct substitution (think about it if I have x^3h^2 + 3xh^2 as h–>0 they will become 0, so the only thing that matters is x^(n-1) since it will have no h next to it since it was divided out)
example of power rule
x^2 becomes 2x, x^3 becomes 3x^2
Note: The derivative of a function always lowers the degree by 1
the derviative of sin x is cos x
Proof: (I really don’t feel like proving this… sorry guys just graph the function to see why or use your trig sum and difference formulas to see that d/dx of sin x is cos x)</p>
<p>Finding the slope at points
x^4 at x=-1, x=0, x=1
d/dx[x^4] = 4x^3
4(-1)^3 = -4 (the slope is negative at this point)
4(0) ^3 = 0 (the slope if 0)
4(1)^3 = 4 (the slope if positive)</p>
<p>Finding an Equation of a Tangent-Line
Find the equation of the tangent line when f(x)=x^2 at (-2,4)
d/dx [x^2] = 2x
2(-2) = -4
our new slope is -4
y-y1 = m(x-x1) (point slope form)
y- 4 = -4(x+2)</p>
<p>Constant Mulitple Rule
(I will give you proofs if you want just pm me,but most people I talked to from cc said that proofs are useless, so I only gave a few proofs from the beginning)
d/dx[cf(x)] = c f’(x)
example y =2x^3
y’ = 2 * d/dx(x^3)
y’ = 2* 3x^2
y’= 6x^2</p>
<p>Sum and difference rules
d/dx[f(x) + or - g(x)] = f’(x) + or - g’(x)
Example y=x^3 -4x +5
y’ = 3x^2+4
Product rule:(pm me for proofs if needed)
f(x)g’(x)+f’(x)g(x)
f(x)= (3x-2x^2)(5+4x)
(d/dx) (3x-2x^2)(5+4x)+ (3x-2x^2) (d/dx) (5+4x)
f’(x)(3-4x)(5+4x) + (3x-2x^2)(4)</p>
<p>Constant Multiple rule vs. product rule
y=(x)^(1/2)g(x) you must use the product rule these are 2 separate functions there is no constant
y=(2)^(1/2) g(x) if you better easier and faster to use the constant multiple rule since you have a constant. However, the product rule still works</p>
<p>Quotient rule (ask for proofs if needed)
(g(x) f’(x) – f(x) g’(x)) / (g(x)) ^2
y = (5x-2)/(x^2+1)
dy/dx = ((x^2+1)(5) – (5x-2)(2x)) /( (x^2+1)^2)
you don’t need to simplify
On the AP exam, if this was a free response question, this would be your final answer. But if this is a multiple choice question, then you might need to simplify </p>
<p>Rewriting before differentiating</p>
<p>find an equation of a tangent line at (-1,1) for f(x) = (3-(1/x))/(x+5)</p>
<p>1st write the original function f(x) = 3 – (1/x)
x+5
f(x) = 3 – (1/x) * x
x+5 *x
f(x) = 3x – 1
x^2+5x<br>
f’(x) = ((x^2+5x)(3) – (3x-1)(2x+5)) / (x^2+5x)^2</p>
<p>f’(-1) = 0 if you sub in</p>
<p>Constant Multiple Rule vs. Power Rule
y = x^2+3x it is easier to use constant multiple because this function is really
6<br>
y = (1/6) (x^2+3x)
However, you can still use quotient rule</p>
<p>Higher order derivatives
s(t) = position function
s’(t) = v(t) velocity function 1st derivative
s’’(t) = v’(t) = a(t) acceleration function 2nd derivative</p>
<p>1st derivative y’ f’(x) dy/dx’ d/dx[f(x)] Dx[y]
2nd derivative y’’ f’’(x) d^2y/dx^2’ d^2/dx^2[f(x)] Dx^2[y]
3rd derivative y’’’ f’’’(x) d^3y/dx^3’ d^3/dx^3[f(x)] Dx^3[y]
4th derivative y^(4) f^4(x) d^4y/dx^4’ d^4/dx^4[f(x)] Dx^4[y]
nth derivative y^n f^n(x) d^ny/dx^5’ d^n/dx^n[f(x)] Dx^n[y]</p>
<p>s(t) = 0.81t^2+2
v(t) = 1.62t
a(t) = 1.62</p>
<p>Chain rule dy/dx =dy/du * du/dx</p>
<p>chain rule = derivate inner function * derivate outer function (keep inner function)</p>
<p>This is for composition of functions
y= (x^2+1)^3
dy/dx = 2x*3(x^2+1)^2
derivate inner function d/dx [(x^2+1)] = 2x
keep inner (x^2+1)
derivate outer d/dx [x^3] = 3x^2
(x^2+1) becomes your x 3(x^2+1)^2</p>
<p>Chain and power
f(x) = (3x-2x^2) ^3
f’(x) = (3-4x) * 3(3x-2x^2)^2
d/dx inner * d/dx out(keep inner)</p>
<p>Btw with chain rule you never have to do quotient rule watch
y = 7x
(x+2)
7x * (x+2) ^-1 then you would use product and chain rule
2
y = [(3x-1) ] </p>
<hr>
<pre><code> [(x^2+3)]
</code></pre>
<p>(the whole quantity is squared)
chain rule : d/dx (3x-1) * 2[(3x-1) ] ^1
(x^2+3) [(x^2+3)]<br>
Quotient rule
3(x^2) – (2x)(3x-1) * 2[(3x-1) ] ^1 (prentend that there is a divison bar)
(x^2+3)^2 [(x^2+3)] </p>
<p>note whenever you see like 2x-3 pretend that there is a divison bar between them
3x-2</p>
<p>Thanks Again CscMath Asa very useful. I think I may be adding that one to my notebook for now!</p>
<p>CSMathAsa1994 - I worship you</p>
<p>Remember guys if you want proofs just pm me. (My teacher said that proofs will not be on the AP exam… but if u plan on doing well in calc 2 in college and other higher level maths, then you might need to know the proofs)
proofs for the intro to derivatives unit include:
power rule
constant multiple rule
sum and difference rule
product rule
quotient rule
chain rule</p>
<p>Got 100 on my limits test because of your guide and now we’re on derivatives and differentiability and such so I’ll most definitely be checking your guide out. CSMath, you’re literally the greatest person on the planet</p>
<p>I got a 100 on my limits test too!!! Thanks guys for calling my guide great. But I would like to know if my guide is overkill for your tests, just right for your tests, or not enough.</p>
<p>Was anyone else’s Precalc class last year a very poor preparation for Calc? I mean, I had an absolutely fantastic teacher and learned quite a bit, but all different topics than this year. A more apt name for “Pre-calculus” would be “Post-trigonometry” in my opinion ;)</p>
<p>I would kinda agree with the post-trigonometry part. My precal class last year didn’t start any like calculus stuff until the last month or 2 of the school year. never really thought about that til now actually.
Also, just a general question, what prep books are you guys using with this class if any? For my AP spanish (even though the teacher gave us Barron’s for free…) and AP physics classes I already see people walking around with their princeton reviews and I thought it was kinda early but I guess they use it to supplement the class</p>
<p>My Pre-cal class sucked. We used the CPM pre-cal textbooks … ugh …</p>
<p>Ok so quiz on derivatives tomorrow and I have a ton of questions…
Ok so really the main one I don’t get at all is when they give you an equation and they tell you to find where its not differentiable. Like for example, my teacher gave us the equation (x+2)^1/3 - 1 and I wasn’t differentiable at x= -2. Help please!
I have more if anyone wants to help…</p>
<p>There is a horizontal tangent line @ x = -2. Thus, it is not differentiable; the slope does not exist/is not a real number.</p>