anonymous
  • anonymous
Problem : Find the derivatives of the function using the limit of definition y=x^4
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
still the same type of question? try going http://www.math.hmc.edu/calculus/tutorials/limit_definition/ they show you examples. just apply your question to how they solve it.
anonymous
  • anonymous
It cant be that you were taught 2 other similar questions but still cant solve it.
anonymous
  • anonymous
im confused about the exponent 4

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anonymous
  • anonymous
do you know how to do (a+b)^4?
anonymous
  • anonymous
no
anonymous
  • anonymous
its the binomial theorem or pascal triangle
anonymous
  • anonymous
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 etc....
anonymous
  • anonymous
for example just look at the pascal triangle for the coefficients (a+b)^2= 1a^2 + 2ab +1b^2= a^2 +2ab+b^2 can you guess whats (a+b)^3? and for (a+b)^4?
anonymous
  • anonymous
note it can be like (x+y)^3 ? or (x+y)^4 ?
anonymous
  • anonymous
so power of 3 we get the coefficients of 1 3 3 1 therefore the answer for (x+y)^3= x^3 +3x^2y +3 xy^2 + y^3 can you guess for (x+y)^4 ?
anonymous
  • anonymous
the reason i want you to learn this is when you go solved for the derivative of y=x^4 you will get to find (x+y)^4
anonymous
  • anonymous
tnx a lot . i need to study more about this
anonymous
  • anonymous
im here now to teach you the easier way of doing it
anonymous
  • anonymous
ok for the power of 4 the coefficents inthe pascal triangle are 1 4 6 4 1 therefore (a+b)^4 = a^4 + 4a^3 b +6 a^2 b^2 +4 a b^3 + b^4 or (x+y)^4= x^4 +4x^3 y + 6 x^2 y^2 + 4x y^3 + y^4 did you see it ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
now for practice y=x^2 \[y+\Delta y=x +\Delta x\] \[\Delta y=x+\Delta x - y\] \[\Delta y= (x+\Delta x)^{2}-x ^{2}\]
anonymous
  • anonymous
can you try solving that first ?
anonymous
  • anonymous
wait
anonymous
  • anonymous
i try solving it
anonymous
  • anonymous
the answer is 2x
anonymous
  • anonymous
\[\Delta y=f(x+\Delta x) - y\] \[\Delta y=(x+\Delta x)^{2}- x ^{2}\] \[\Delta y= x ^{2 }+ 2x \Delta x +\Delta x ^{2} -x ^{2}\] add or subtract we get \[\Delta y= 2x \Delta x+\Delta x ^{2}\] now the derivative is \[y'=f'(x)= \lim _{\Delta x ->0}\frac{ \Delta y }{ \Delta x }=\] \[= \lim _{\Delta x->0}(\frac{ 2x \Delta x+\Delta x ^{2} }{ \Delta x })\] try to cancell something in there
anonymous
  • anonymous
i know it already the answer is 2x
anonymous
  • anonymous
ok good can you try it if its y=x^4 ? do the same process
anonymous
  • anonymous
okay
anonymous
  • anonymous
the answer is 4x^3+6x^2+4x
anonymous
  • anonymous
hmm to make it easier lets make delta x=h so that delta y=(x+h)^4 - x^4 try solving that first
anonymous
  • anonymous
I cant follow my professor's teaching. Good to know that u are here to help me .thank you so much. you are a great´╗┐ help! i need to do more assignments about this so i will go out now .
anonymous
  • anonymous
the answer should be f'(x)=4x^3
anonymous
  • anonymous
haha im wrong
anonymous
  • anonymous
do the process i gave you there, and you will arrive at the correct answer ... :D
anonymous
  • anonymous
i wish i have a tutor like u hahaha .
anonymous
  • anonymous
remember this ?. (x+y)^4= x^4 +4x^3 y + 6 x^2 y^2 + 4x y^3 + y^4 make y an h here so that (x+h)^4= x^4 +4x^3 h + 6 x^2 h^2 + 4x h^3 + h^4
anonymous
  • anonymous
anyway thx a lot
anonymous
  • anonymous
goodbye!
anonymous
  • anonymous
ok just read it through here again and learn from it good luck and have fun now :D
anonymous
  • anonymous
ok
anonymous
  • anonymous
ok just messege me here and if you have prob i will go and give you an advise or correct if you have something done incorrectly ... :D

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