Problem :
Find the derivatives of the function using the limit of definition
y=x^4

- anonymous

Problem :
Find the derivatives of the function using the limit of definition
y=x^4

- Stacey Warren - Expert brainly.com

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- jamiebookeater

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- 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

It cant be that you were taught 2 other similar questions but still cant solve it.

- anonymous

im confused about the exponent 4

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## More answers

- anonymous

do you know how to do
(a+b)^4?

- anonymous

no

- anonymous

its the binomial theorem or pascal triangle

- anonymous

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
etc....

- 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

note it can be like (x+y)^3 ?
or (x+y)^4 ?

- 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

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

tnx a lot . i need to study more about this

- anonymous

im here now to teach you the easier way of doing it

- 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

yes

- 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

can you try solving that first ?

- anonymous

wait

- anonymous

i try solving it

- anonymous

the answer is 2x

- 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

i know it already the answer is 2x

- anonymous

ok good can you try it if its y=x^4 ? do the same process

- anonymous

okay

- anonymous

the answer is 4x^3+6x^2+4x

- 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

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

the answer should be f'(x)=4x^3

- anonymous

haha im wrong

- anonymous

do the process i gave you there, and you will arrive at the correct answer ... :D

- anonymous

i wish i have a tutor like u hahaha .

- 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

anyway thx a lot

- anonymous

goodbye!

- anonymous

ok just read it through here again and learn from it
good luck and have fun now :D

- anonymous

ok

- 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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