Solve Derivative of Function at a Specific Value for x w/ First Principles

[Petrus]

Hello!
I am stuck trying to understanding this question. I would be glad if someone could explain it to me.

"Each limit represents the derivative of some function $f$ at some number $a$. State such an $f$ and $a$ in each case."

Example of a problem:

$\displaystyle \lim_{h\to0}\frac{(1+h)^{10}-1}{h}$

 

[MarkFL]

Re: Limit,derivate

If I am understanding correctly, you wish to use first principles to find the derivative of:

$f(x)=x^{10}$

at $x=1$

Hint: Use the binomial theorem. This theorem can be used to prove the power rule for integral exponents. Can you state the problem using this theorem?

 

[alyafey22]

Re: Using first principles to find the derivative of a function at a specific vlaue for x

To simplify computations , first use difference of two squares rule.

 

[Petrus]

Re: Limit,derivate

If I am understanding correctly, you wish to use first principles to find the derivative of:

$f(x)=x^{10}$

at $x=1$

Hint: Use the binomial theorem. This theorem can be used to prove the power rule for integral exponents. Can you state the problem using this theorem?

Hello Mark!
Thank you for the help:) On this chapter, we "learn" to use lim h->0(f(a+h)-f(a))/h. About binomal I don't know how to do that.
edit: ok so I don't know if i done right but here is what i did
first i use the binomal on (1+h)^10 and get 1^10+10•1^9•h+90/2•1^8•h^2 and simplifie and get answer 10^9 while the facit says x^10, a=1

 

[MarkFL]

Re: Using first principles to find the derivative of a function at a specific vlaue for x

Hey Petrus,

We ask that you do not use text message abbreviations, since many students who may read this may not be familiar with them and would find your posts hard to follow. I have edited your post to write them out fully.

Now, the binomial theorem states:

$\displaystyle (x+y)^n=\sum_{k=0}^n{n \choose k}x^{n-k}y^{k}$

Can you see how all but the first term in the expansion of $(1+h)^{10}$ will contain $h$ as a factor, and this first term will be subtracted away, so that the numerator will have a series of terms with $h$ as a factor, that you can then divide out with the $h$ in the denominator? Once this is done, you will be left with but one term that will not vanish as $h\to0$.

Can you write this out?

edit: I must turn in now, so anyone who wishes to reply to further posts may do so.

 

[Petrus]

Re: Using first principles to find the derivative of a function at a specific vlaue for x

Hey Petrus,

We ask that you do not use text message abbreviations, since many students who may read this may not be familiar with them and would find your posts hard to follow. I have edited your post to write them out fully.

Now, the binomial theorem states:

$\displaystyle (x+y)^n=\sum_{k=0}^n{n \choose k}x^{n-k}y^{k}$

Can you see how all but the first term in the expansion of $(1+h)^{10}$ will contain $h$ as a factor, and this first term will be subtracted away, so that the numerator will have a series of terms with $h$ as a factor, that you can then divide out with the $h$ in the denominator? Once this is done, you will be left with but one term that will not vanish as $h\to0$.

Can you write this out?

edit: I must turn in now, so anyone who wishes to reply to further posts may do so.

Hello Mark!
i unfortenetly i Edit My post with the method that is same as that. I Will lean this latex later because the way i use to write it does not work:/ i Will learn soon as possible I am on pc its diffrent from \(\displaystyle . Any1 could check My Edit i made on My post early ty.\)

 

[I like Serena]

Re: Using first principles to find the derivative of a function at a specific vlaue for x

Welcome to MHB, Petrus! :)

Hello Mark!
i unfortenetly i Edit My post with the method that is same as that. I Will lean this latex later because the way i use to write it does not work:/ i Will learn soon as possible I am on pc its diffrent from \(\displaystyle . Any1 could check My Edit i made on My post early ty.\)

\(\displaystyle

I believe Mark already gave the answer to your problem in post #2.
It seems you are not supposed to evaluate the limit, so you do not really need the binomial theorem.

To clarify, if you substutite $f(x)=x^{10}$ into:

$\qquad\displaystyle\lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h}$

you get:

$\qquad\displaystyle\lim_{h \to 0} \dfrac{(a+h)^{10} - (a)^{10}}{h}$

And if you now substitute $a=1$, you get:

$\qquad\displaystyle\lim_{h \to 0} \dfrac{(1+h)^{10} - (1)^{10}}{h}$

This is the formula you were after, so the answer is $f(x)=x^{10}$ and $a=1$.

Btw, can you please please write regular English?
Trying to read and understand what you write makes my head hurt.\)

 

[MarkFL]

Re: Using first principles to find the derivative of a function at a specific vlaue for x

English is not the native language of Petrus (as you can see, he is from Sweden), so we should forgive spelling and grammatical errors, however, things like "ty" in place of "thank you" or "any1" in place of "anyone" or "idk" in place of "I don't know" should be avoided for the reason I stated above.

I also know that Petrus is not familiar with $\LaTeX$ so we should forgive this for a spell as he learns it. Petrus, we have a "LateX Help" forum that contains tips on its usage, and you may post any questions you have about getting started with it there.

I apologize Petrus for making the problem more difficult than it needed to be. I figured you also needed to evaluate the limit, and thought identifying $f(x)$ and $a$ was not where you were stuck.

 

[Chris L T521]

Re: Using first principles to find the derivative of a function at a specific vlaue for x

Hello!
I am stuck trying to understanding this question. I would be glad if someone could explain it to me.

"Each limit represents the derivative of some function $f$ at some number $a$. State such an $f$ and $a$ in each case."

Example of a problem:

$\displaystyle \lim_{h\to0}\frac{(1+h)^{10}-1}{h}$

After typing this out, I realized I misunderstood the problem, but if you don't mind, I'd like to add my two cents on another way to compute the derivative of this guy from first principles.

There is another way to find the derivative of $x^{10}$ from first principles without using the binomial theorem as MarkFL originally suggested, but in the end requires you to notice a peculiar form of factorizing a certain expression. In addition to the well known formula $f^{\prime}(a)=\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$, we also have an alternate (yet equivalent) form $f^{\prime}(a)=\displaystyle\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$, which I personally like to use when having to justify the derivatives of polynomials from first principles.

So if we wanted to compute the derivative of $x^{10}$ at $x=1$, we'd have $f^{\prime}(1)=\displaystyle\lim_{x\to 1}\frac{x^{10}-1}{x-1}$. So now, there's actually a way to factorize the above expression. To see this, let us factor a few expressions and notice the pattern.

We know that:

$\hspace{.5in}\begin{aligned} x^2 - 1 &= (x-1)(x+1)\\ x^3-1 &= (x-1)(x^2+x+1)\\ x^4-1 &= (x^2-1)(x^2+1)\\ &= (x-1)(x+1)(x^2+1)\\ &= (x-1)(x^3+x^2+x+1)\end{aligned}$

So following the same pattern, we'd observe that $x^{10}-1=(x-1)(x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)$.

(If this was done for general $a$, we'd notice that $x^{10}-a^{10}=(x-a)(x^9+ax^8+a^2x^7+a^3x^6+a^4x^5+a^5x^4+a^6x^3+a^7x^2+a^8x+a^9)$.)

Therefore,
$\hspace{.5in}\begin{aligned}\lim_{x\to 1}\frac{x^{10}-1}{x-1} &= \lim_{x\to 1}\frac{(x-1)(x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)}{x-1}\\ &=\lim_{x\to 1}x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1\\ &= 1+1+1+1+1+1+1+1+1+1\\ &= 10\end{aligned}$

Thus, for $f(x)=x^{10}$, $f^{\prime}(1)=10$.

I hope this makes sense!

 

[I like Serena]

Re: Using first principles to find the derivative of a function at a specific vlaue for x

As long as we're at it, let me also add my 2 cents.

The expression $\dfrac{x^{10}-1}{x-1}$ is a quotient that we want to simplify.
Usually a quotient is simplified with a long division.
In particular that also works for polynomials:

$
\begin{array}{ll}
& \underline{x^9 + x^8 + x^7 + ... + x + 1} \\
x-1 ) & x^{10} \phantom{+ x^8 + x^7 + ... + x + 1} - 1 \\
&\underline{x^{10} - x^9} \\
&\phantom{x^{10} +} x^9 \\
&\phantom{x^{10} +} \underline{x^9 - x^8} \\
&\phantom{x^{10} + x^9 +} x^8 \\
&\phantom{x^{10} + x^9 +} \underline{x^8 - x^7} \\
&\phantom{x^{10} + x^9 + x^8 +} x^7 \\
&\phantom{x^{10} + x^9 + x^8 +} \underline{... \phantom{0}} \\
&\phantom{x^{10} + x^9 + x^8 + ...} 0 \\
\end{array}
$

In other words:

$\dfrac{x^{10}-1}{x-1} = x^9 + x^8 + x^7 + ... + x + 1$

Phew! That was quite a challenge to do it with latex! :)