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blogfast25
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We'll accept green.
The following for today/tomorrow:
Polynomials:
$$y=3x^{-2}-7x^3+x^2-19$$
$$u=ax^3-bx^2+c\:\text{(where a, b and c are constants)}$$
$$z=3x^a-5x^b\:\text{(where a and b are constants)}$$
Product rule exercises:
Worked example:
$$f(x)=(2x-9)(5-x^2)$$
$$f'(x)=(2x-9)'(5-x^2)+(2x-9)(5-x^2)'=2(5-x^2)+(2x-9)(-2x)$$
Further simplified:
$$f'(x)=10-2x^2-4x^2+18x=10+18x-6x^2$$
Exercises:
$$y=x^{-2}(3x-5)$$
$$u=(2x^7+a)(x^6-b)\:\text{(where a and b are constants)}$$
$$z=(x-1)\sin x\:\text({note: (sin x)'=cos x})$$
Quotient rule exercises:
Worked example:
$$f(x)=\frac{2x-a}{3x+a}$$
$$f'(x)=\frac{(2x-a)'(3x+a)-(2x-a)(3x+a)'}{(3x+a)^2}$$
$$f'(x)=\frac{2(3x+a)-(2x-a)3}{(3x+a)^2}$$
$$f'(x)=\frac{5a}{(3x+a)^2}$$
Exercises:
$$y=\frac{\sin x}{x}$$
$$u=\frac{x^3-1}{\sqrt{x}}$$
$$\text{note:}\:\sqrt{x}=x^{\frac12}$$
$$z=\frac{\ln x}{x-5}\:\text{(note: (lnx)'=1/x)}$$
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aga
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OK. Thanks. Will get head-scratching on those.
One bit i cannot fathom from much earlier is
$$\frac{x}{y}=0$$
how can any combination = 0, even with the extra special case of x=y=0 ?
Edit:
... and what is a polynomial actually ? (googled, still unsure)
[Edited on 21-3-2016 by aga]
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aga
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The latex is tricky to get right 1st time (easier on paper).
One by one ...
$$y=3x^{-2}-7x^3+x^2-19$$
$$y'=3(-2x^{-3})-7(3x^2)+x+0$$
$$y'=-6x^{-3}-21x^2+x$$
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aga
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next ...
$$u=ax^3-bx^2+c\:\text{(where a, b and c are constants)}$$
Eh ? if they're constants their derivitives are zero, so ..
$$u'=0$$
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aga
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next ...
$$z=3x^a-5x^b\:\text{(where a and b are constants)}$$
Hmm.
$$z' = 3(ax^{a-1})-5(bx^{b-1})$$
Erm, we're done processing the derivative rules, so i guess that's the answer. Maybe expand the brackets :-
$$z'=3ax^{a-1}-5bx^{b-1}$$
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aga
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next ...
$$y=x^{-2}(3x-5)$$
oooh ! i think the product rule applies
$$y'=(x^{-2})'(3x-5) + (x^{-2})(3x-5)'$$
$$y'=(-2x)(3x-5)+3(x^{-2})$$
$$y'=-6x^2+10x+3x^{-2}$$
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aga
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next ...
$$u=(2x^7+a)(x^6-b)\:\text{(where a and b are constants)}$$
product rule again ...
$$u' = (2x^7+a)'(x^6-b)+(2x^7+a)(x^6-b)'$$
so ...
$$u' = (2(7x^6) +0)(x^6-b)+(2x^7+a)(6(x^5)-0)$$
$$u' = (14x^6)(x^6-b)+(2x^7+a)(6x^5)$$
multiply out ...
$$u'=14x^{12}-14bx^6 + 12x^{12} +6ax^5$$
simplify...
$$u'=26x^{12}-14bx^6+6ax^5$$
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aga
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i do hope i get to keep my snotty hanky, at least for re-arranging the answers so the biggest exponent comes first.
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aga
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next ...
$$z=(x-1)\sin x\:\text({note: (sin x)'=cos x})$$
Erm, product rule again, maybe ...
$$z'=(x-1)'(sin x)+(x-1)(sin x)'$$
$$z'=sin x + cos x (x-1)$$
$$z'=sin(x) +x cos(x)-cos(x)$$
No idea which term would be bigger so i guess i lost the hanky.
The cos(x)-cos(x) and the sin(x) all look kinda familiar.
Is there a rule that says xcos(x)-cos(x) = twelvety ?
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aga
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The next ones look harder.
Here goes ..
$$y=\frac{\sin x}{x}$$
Quotient rule (why they don't call it the Divide-By rule is a mystery)
$$y'=\frac{x sin(x)'-sin(x)x'}{x^2}$$
$$y'=\frac{xcos(x)-sin(x)}{x^2}$$
I got a bad feeling about this one.
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woelen
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Your results are quite good, but there are some calculation errors. I do not think these are due to lack of understanding, but due to being somewhat
imprecise. With these calculus-things you have to be very precise.
Where things are wrong I will put a response to that post.
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woelen
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Quote: Originally posted by aga  | The latex is tricky to get right 1st time (easier on paper).
One by one ...
$$y=3x^{-2}-7x^3+x^2-19$$
$$y'=3(-2x^{-3})-7(3x^2)+x+0$$
$$y'=-6x^{-3}-21x^2+x$$
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The more complicated term is OK, but you made a mistake with the simple x2. The derivative is not x, but 2x. The total answer should be
$$y'=-6x^{-3}-21x^2+2x$$
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woelen
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| Quote: Originally posted by aga | next ...
$$u=ax^3-bx^2+c\:\text{(where a, b and c are constants)}$$
Eh ? if they're constants their derivitives are zero, so ..
$$u'=0$$
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The numbers a and b are constants, but they are multiplied with powers of variable x. Using the power rule you get
$$3ax^2 - 2bx$$
The constant c indeed leads to value 0, hence it disappears from the final result.
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woelen
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| Quote: Originally posted by aga | next ...
$$y=x^{-2}(3x-5)$$
oooh ! i think the product rule applies
$$y'=(x^{-2})'(3x-5) + (x^{-2})(3x-5)'$$
$$y'=(-2x)(3x-5)+3(x^{-2})$$
$$y'=-6x^2+10x+3x^{-2}$$ |
The derivative of x-2 is equal to -2x-2-1 = -2x-3.
You did the correct factor -2, but for the power you did as if it is +2 and hence you end up with -2x for the derivative of x-2. This seems
to me like understanding the rules, but not being sufficiently precise/meticulous.
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blogfast25
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@aga:
Not sure where you're coming from with:
$$\frac{x}{y}=0$$
That would be true for:
$$x=0\:\text{OR:}y=\infty$$
But I don't see where I brought it up?
Perhaps you meant:
$$\frac{\Delta x}{\Delta y}$$
A polynomial: a sum of terms of the general form:
$$ax^2$$
E.g.:
$$y=2+3x-x^5$$
... is a polynomial.
Exercises:
$$y'=3(-2x^{-3})-7(3x^2)+x+0$$
... is a very minor error, should have been:
$$y'=3(-2x^{-3})-7(3x^2)+2x+0$$
The next one was wrong. a, b and c are constants but u is definitely a function of x and its derivative is:
$$u'=3ax^2-2bx$$
Next:
$$z'=3ax^{a-1}-5bx^{b-1}$$
... is 100 % correct.
Next:
$$y'=-6x^2+10x+3x^{-2}$$
... see woelen
Next:
$$u'=26x^{12}-14bx^6+6ax^5$$
... is 100 % correct. Well done also on simplifying it.
Next:
$$z'=\sin(x) +x \cos(x)-\cos(x)$$
... is 100 % correct.
<hr>
Looks like you're getting the hang of it.
Your prize: a bit of reminder algebra of powers:  just in case.
$$x^nx^m=x^{n+m}$$
$$\frac{x^n}{x^m}=x^{n-m}$$
$$\Big(x^n\Big)^m=x^{nm}$$
$$x^0=1$$
$$\frac{1}{x^n}=x^{-n}$$
$$\sqrt{x}=x^\frac12$$
$$\sqrt[n]{x}=x^\frac1n$$
[Edited on 21-3-2016 by blogfast25]
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aga
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Many thanks for taking the time to analyse and correct the errors woelen.
It is very much appreciated.
Some innacurracies can be attributed to being a Noob to calculus.
Yes, any slight error can easily throw the final result into chaos, so meticulousness is required for this.
Likely that the majority of the errors can be put down to the fact that it's almost half-past midnight here and i'm not exacty sober.
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aga
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Erm., bloggers, you do realiase that by posting those 'reminders' that you're giving away the secrets of the illuminati (which i print out and guard
jealously) and that they might try to stop you one day ?
Woohoo ! I got some right !
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blogfast25
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| Quote: Originally posted by aga | The next ones look harder.
Here goes ..
$$y=\frac{\sin x}{x}$$
Quotient rule (why they don't call it the Divide-By rule is a mystery)
$$y'=\frac{x sin(x)'-sin(x)x'}{x^2}$$
$$y'=\frac{xcos(x)-sin(x)}{x^2}$$
I got a bad feeling about this one. |
... is correct.
[Edited on 21-3-2016 by blogfast25]
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aga
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That's 4 out of 38 which is 100% !
Woohoo !
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blogfast25
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Give or take:
$$\sqrt{c-5}$$
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blogfast25
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Tomorrow:
Simple chain rule exercises. Fun and games!
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blogfast25
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This exercise got a bit overlooked, so I'll just post the solution here:
$$z=\frac{\ln x}{x-5}\:\text{(note: (lnx)'=1/x)}$$
Solution:
$$z'=\frac{(\ln x)'(x-5)-\ln x (x-5)'}{(x-5)^2}$$
$$z'=\frac{\frac1x (x-5)-\ln x}{(x-5)^2}$$
Simplified:
$$z=\frac{x-x\ln x-5}{x(x-5)^2}$$
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blogfast25
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1. Chain rule exercises:
Recapping the rule. If:
$$f(u)$$
with u a function of x:
$$u(x)$$
Then:
$$f'(x)=f'(u) \times u'$$
Worked example:
$$f(x)=\sqrt{3x^2+5x-8}$$
$$f'(x)=\frac12(3x^2+5x-8)^{-\frac12}(3x^2+5x-8)'$$
$$f'(x)=\frac{6x+5}{2\sqrt{3x^2+5x-8}}$$
Simple exercises:
$$y=\sin (2x^3)$$
$$u=(1+\cos x)^4$$
$$v=\Big(a+\frac1x\Big)^2$$
$$z=\ln (x^5+1)$$
2. Chain rule exercises with exponential functions:
If, with e = 2.71828182845904523536028747135266249775724709369995... :
$$f(x)=e^x$$
Then:
$$f'(x)=e^x$$
(Yep, it's that simple: the derivative of the function is the function itself!)
Now, if:
$$f(x)=e^{u(x)}$$
With u a function of x, then:
$$f'(x)=e^{u(x)} \times u'(x)$$
Example:
$$f(x)=e^{3x}$$
$$f'(x)=e^{3x}(3x)'=3e^{3x}$$
Exercises:
$$y=e^{x+1}$$
$$u=e^{x^2-3}$$
$$v=e^{\sin x}$$
$$z=e^{\sqrt{2-x^3}}$$
$$f=e^{0.5231x}$$
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aga
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Not sure where to go with these chain rule exercises.
Here's what i imagine needs doing for the first one :-
$$y=sin (2x^3)$$
$$y'=[sin(2x^3)]'(2x^3)'$$
$$y'=cos(2x^3).6x^2$$
The y' bit seems wrong as does the application of the rule.
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aga
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I forgot that all the Masters of the B&D Uni go on the traditional bar and brothel crawl during Easter Week.
Oh well, perhaps these will work out OK.
If not it'll be rubbing linseed oil into the school Cormorant for a week.
next
$$u=(1+cos x)^4$$
$$u'=[(1+cosx)^4]'.(1+cosx)'$$
$$u'=4(1+cosx)^3.-sinx$$
next (presuming a is a constant)
$$v=\Big(a+\frac1x\Big)^2$$
$$v'=\Big[(a+\frac1x)^2]'\Big.(a+\frac1x)'$$
$$v'=2\Big(a+\frac1x)\Big.-x^{-2}$$
next
$$z=\ln (x^5+1)$$
$$z'=[ln(x^5+1)]'.(x^5+1)'$$
$$z'=\frac{5x^4}{x^5+1}$$
next
$$y=e^{x+1}$$
$$y'=e^{x+1}(x+1)'$$
$$y'=e^{x+1}$$
next
$$u=e^{x^2-3}$$
$$u'=e^{x^2-3}.(x^2-3)'$$
$$u'=e^{x^2-3}.2x$$
next
$$v=e^{sin x}$$
$$v'=e^{sinx}(sinx)'$$
$$v'=e^{sinx}.cosx$$
next
$$z=e^{\sqrt{2-x^3}}$$
$$z'=e^{\sqrt{2-x^3}}.(\sqrt{2-x^3})'$$
$$z'=e^{\sqrt{2-x^3}}.((2-x^3)^{\frac12})'$$
$$z'=e^{\sqrt{2-x^3}}.\frac12(2-x^3)^{-\frac12}$$
finally
$$f=e^{0.5231x}$$
$$f'=e^{0.5231x}.(0.5231x)'$$
$$f'=e^{0.5231x}.0$$
$$f'=0$$
[Edited on 23-3-2016 by aga]
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