Algebraic Function Derivatives

[Jason76]

Basically I get the gist of what's going on. But a few problems remain.

The power rule and chain rule applies to these derivative problems:

1.

\(\displaystyle y = (2x^{3} - 5x^{2} + 4)^{5}\)

Utilize Power Rule:

First, turn the 5 exponent (which is outside the parenthesis) into a 5 coefficient (in front of the parenthesis). However, keep an exponent outside the parenthesis and make it 4.

\(\displaystyle 5(2x^{3} - 5x^{2} + 4)^{4}\)

Utilize Chain Rule:

Now make another linear factor to the right of the problem, differintiate the numbers in the left linear factor (using Power Rule), put them in the right linear factor:

\(\displaystyle (6x^{2} - 10x)\)

Now put them side by side:

\(\displaystyle 5(2x^{3} - 5x^{2} + 4)^{4} (6x^{2} - 10x) \)

However, it stops here (according to the book). There is no distributive property with the 5 in the left linear factor. Why?

Now let's contrast this problem with problem 2:

2.

\(\displaystyle y = (x^{2} + 2)^{3}\)

Divided into two linear factors (chain rule)

\(\displaystyle 3(x^{2} + 2)^{2} (2x)\) (the 2x coming from differentiating inside of the original problem - "Power Rule")(The power rule is applied as the exponent of 3 is reduced to 2 and 3 is put in front of the left linear factor.)

Now the left linear factor is factored.

\(\displaystyle 3(x^{4} + 4x^{2} + 4) (2x)\)

2x is multiplied with 3 and a distributive property takes place (for the final answer).

\(\displaystyle 6x(x^{4} + 4x^{2} + 4)\)

\(\displaystyle 6x^{5} + 24x^{3} + 24x\)

This answer had several steps the first one did not have.

Let's look at one final 3rd problem:

3.

\(\displaystyle y = (x^{3} - 4)^{5}\) (Some explanations of Power Rule or Chain rule ommited)

No factoring takes place in the left linear factor. However, since we have a single number in the right linear factor with an x (or single number with an x and some exponent of just an x with an exponent), then we multiply the number in the right linear factor with 5.


\(\displaystyle 5(x^{3} - 4)^{4}(3x^{2})\)


\(\displaystyle 15x^{2}(x^{3} - 4)^{4}\) (Final Answer)

 

[tkhunny]

Why are you troubled by algebraic manipulation? The general idea is to simplify most expresssions to make them more convenient. Do you REALLY want to multiply out a trinomial raised to the 4th power? You should probably try to work on the purpose, rather than the rote memorization of some mechanical process.

 

[Jason76]

Why are you troubled by algebraic manipulation? The general idea is to simplify most expresssions to make them more convenient. Do you REALLY want to multiply out a trinomial raised to the 4th power? You should probably try to work on the purpose, rather than the rote memorization of some mechanical process.

I can see why, in some cases, factoring is NOT done. However, the 2nd problem used the distributive property, while the other ones did not. I didn't understand why.

 

[JeffM]

Basically I get the gist of what's going on. But a few problems remain.

The power rule and chain rule applies to these derivative problems:

1.

\(\displaystyle y = (2x^{3} - 5x^{2} + 4)^{5}\)

Utilize Power Rule:

First, turn the 5 exponent (which is outside the parenthesis) into a 5 coefficient (in front of the parenthesis). However, keep an exponent outside the parenthesis and make it 4.

\(\displaystyle 5(2x^{3} - 5x^{2} + 4)^{4}\)

Utilize Chain Rule:

Now make another linear factor to the right of the problem, differintiate the numbers in the left linear factor (using Power Rule), put them in the right linear factor:

\(\displaystyle (6x^{2} - 10x)\)

Now put them side by side:

\(\displaystyle 5(2x^{3} - 5x^{2} + 4)^{4} (6x^{2} - 10x) \)

However, it stops here (according to the book). There is no distributive property with the 5 in the left linear factor. Why?

Now let's contrast this problem with problem 2:

2.

\(\displaystyle y = (x^{2} + 2)^{3}\)

Divided into two linear factors (chain rule)

\(\displaystyle 3(x^{2} + 2)^{2} (2x)\) (the 2x coming from differentiating inside of the original problem - "Power Rule")(The power rule is applied as the exponent of 3 is reduced to 2 and 3 is put in front of the left linear factor.)

Now the left linear factor is factored.

\(\displaystyle 3(x^{4} + 4x^{2} + 4) (2x)\)

2x is multiplied with 3 and a distributive property takes place (for the final answer).

\(\displaystyle 6x(x^{4} + 4x^{2} + 4)\)

\(\displaystyle 6x^{5} + 24x^{3} + 24x\)

This answer had several steps the first one did not have.

Let's look at one final 3rd problem:

3.

\(\displaystyle y = (x^{3} - 4)^{5}\) (Some explanations of Power Rule or Chain rule ommited)

\(\displaystyle 5(x^{3} - 4)^{4}(3x^{2})\)

No factoring takes place in the left linear factor. However, since we have a number in the right linear factor with an x (or number with an x and some exponent of just an x with an exponent), then we multiply the number in the right linear factor with 5.

\(\displaystyle 15x^{2}(x^{3} - 4)^{4}\) (Final Answer)

The simplest way that I know how to explain the chain rule is with substitution. Let's take the first problem as an example.

\(\displaystyle Let\ a = 2x^3 - 5x^2 + 4 \implies \dfrac{da}{dx} = 6x^2 - 10x.\)

We used the power and addition rules. Now we substitute.

\(\displaystyle And\ y = (2x^3 - 5x^2 - 4)^5 \implies y = a^5 \implies \dfrac{dy}{da} = 5a^4.\)

We used the power rule.

But we want \(\displaystyle \dfrac{dy}{dx}\), not \(\displaystyle \dfrac{dy}{da}\).

Now the chain rule comes into play.

\(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{da} * \dfrac{da}{dx} = 5a^4 * (6x^2 - 10x) = 5(2x^3 - 5x^2 - 4)^4(6x^2 - 10x).\)

If you want to distribute the 5, you may do so over \(\displaystyle (6x^2 - 10x)\).

You may also do so over the other term AFTER you raise it to the fourth power.

 

[mmm4444bot]

Is the "book" again that questionable web site? Maybe you are paraphrasing? Some very sloppy terminology in that instruction. :-?

 

[lookagain]

:

1.

\(\displaystyle y = (2x^{3} - 5x^{2} + 4)^{5}\)

\(\displaystyle y' \ = \ 10x(3x - 5)(2x^3 - 5x^2 + 4)^4\) <---Simplified


It is true that \(\displaystyle (2x^3 - 5x^2 + 4) = (x - 2)(2x^2 - x - 2)\)


But then \(\displaystyle \ \ y' \ = \ 10x(3x - 5)\bigg((x - 2)(2x^2 - x - 2)\bigg)^4, \ \ \ or\)


\(\displaystyle y' \ = \ 10x(3x - 5)(x - 2)^4(2x^2 - x - 2)^4\)


But then this needlessly increases the number of factors.


2.

\(\displaystyle y \ = \ (x^{2} + 2)^{3}\)

\(\displaystyle y' \ = \ 6x(x^2 + 2)^2\) <--- Simplified.




3.

\(\displaystyle y \ = \ (x^{3} - 4)^{5}\)

\(\displaystyle y' \ = \ 15x^{2}(x^{3} - 4)^{4}\) (Final Answer) <---Yes, this is simplified.

By the way, a complete answer includes the "\(\displaystyle y' = \)" part or the "\(\displaystyle \dfrac{dy}{dx} \ = \)" part.



*** Edit *** "Simplified is in the eye of the grader."


And certain graders are cut-and-dried wrong about what they claim to be "simplified."

 

[Deleted member 4993]

Like I said before - simplified is in the eye of the grader.

 

[Deleted member 4993]

In the bad-old days - days before the graphing calculator - I found advantage in factorizing a function completely (including complex and/or irrational roots). That helped me to visualize the polynomial without actually graphing it.

 

[Jason76]

Is the "book" again that questionable web site? Maybe you are paraphrasing? Some very sloppy terminology in that instruction.

My own words, but new to this stuff. The book is a new current book from the bookstore.

 

[lookagain]

Like I said before - simplified is in the eye of the grader.

But certain graders are outright wrong on certain instructions given on
certain problems at (in the context of) certain mathematics levels.


Example:

The instructions by a teacher in an elementary algebra class state,

"Simplify** the following algebraic expression:


\(\displaystyle 2x^1 - 3x^1 + 3 - 3\)



For instance, if the grader accepts any of the following answers
(but not exclusively these choices), then the grader is wrong:


\(\displaystyle 2x^1 - 3x^1 + 0\)

\(\displaystyle -3x^1 + 2x^1 + 0\)

\(\displaystyle -3x + 2x + 0\)

\(\displaystyle 2x - 3x + 0\)

\(\displaystyle 0 - 3x^1 + 2x^1\)

\(\displaystyle 0 + 2x^1 - 3x^1\)

\(\displaystyle 0 + 2x - 3x\)

\(\displaystyle 0 - 3x + 2x\)

\(\displaystyle 0 - 1x^1\)

\(\displaystyle 0 - x^1\)

\(\displaystyle 0 - 1x\)

\(\displaystyle 0 - x\)

\(\displaystyle -1x^1 + 3 - 3\)

\(\displaystyle -1x + 3 - 3\)

\(\displaystyle -x + 3 - 3\)

\(\displaystyle -x^1 + 3 - 3\)

\(\displaystyle -1x^1 + 0\)

\(\displaystyle -1x + 0\)

\(\displaystyle -x^1 + 0\)

\(\displaystyle -x + 0\)




** This should implicitly mean "simplify to the fullest extent possible."

 

[Jason76]

More Calculus Derivatives

Not sure if this constituted a new thread, so put it on this one:

I understand the following:

1.)

\(\displaystyle h(x) = sqrt/(2x^{3} - 4x + 5)\)

Rewrite radical as an exponent.

\(\displaystyle h(x) = (2x^{3} - 4x + 5)^{1/2}\)

Get the derivatives of the equation and put them in the following right linear factor. Subtract the left linear equation's exponent by 1 and put the old exponent in front of the left linear equation (as a regular number).

\(\displaystyle y = 1/2(2x^{3} - 4x + 5)^{-1/2}(6x^{2} - 4)\)

Now multiply the right linear equation by the number in front of the left linear equation (1/2).

\(\displaystyle y = 1/2(2x^{3} - 4x + 5)^{-1/2}(6x^{2} - 4)\)

That former action leads to this result:

\(\displaystyle y = (3x^{2} - 2)(2x^{3} - 4x + 5)^{-1/2}\)

This could be rewritten as a fraction:

\(\displaystyle y = (3x^{2} - 2)\)

over

\(\displaystyle y = sqrt/(2x^{3} - 4x + 5)\)

because a negative exponent puts an equation or number in the denominator.

However, regarding another problem:

2.)

\(\displaystyle y = sqrt/(3x^{2} + 4)\)

The book's answer isn't what it seems to be leading to.

Books answer:

\(\displaystyle y = 3x(3x^{2} +4)^{1/2}(3x^{2} + 4)^{-1}\)

Which of course, as before, can be written as a fraction with the negative exponated (if that's the right word) linear factor in the bottom. The linear factor with the exponent of positive 1/2 could be re-written as a radical, as we know.

 

[JeffM]

Not sure if this constituted a new thread, so put it on this one:

I understand the following:

\(\displaystyle h(x) = sqrt\(2x^{3} - 4x + 5)\)

Rewrite radical as an exponent.

\(\displaystyle h(x) = (2x^{3} - 4x + 5)^{1/2}\)

Get the derivatives of the equation and put them in the following right linear factor. Subtract the left linear equation's exponent by 1 and put the old exponent in front of the left linear equation (as a regular number).

\(\displaystyle y = 1/2(2x^{3} - 4x + 5)^{-1/2}(6x^{2} - 4)\)

Now multiply the right linear equation by the number in front of the left linear equation (1/2).

\(\displaystyle (3x^{2} - 2)(2x^{3} - 4x + 5)^{-1/2}\)

This could be rewritten as a fraction:

\(\displaystyle (3x^{2} - 2)\)

over

\(\displaystyle sqrt\(2x^{3} - 4x + 5)\)

because a negative exponent puts an equation or number in the denominator.

Yes, you are doing this right, but your description of the process is amazingly cumbersome. I am going to explain why that process works.

\(\displaystyle Given\ h = \sqrt{2x^3 - 4x + 5}.\)

\(\displaystyle Let\ u = 2x^3 - 4x + 5 \implies \dfrac{du}{dx} = 6x^2 - 4.\) That is where your "left hand factor" comes from.

\(\displaystyle Also,\ u = 2x^3 - 4x + 5 \implies h = \sqrt{u} = u^{(1/2)} \implies \dfrac{dh}{du} = \dfrac{1}{2} * u^{-(1/2)}.\) This is the power rule.

You do not need u anymore so \(\displaystyle \dfrac{dh}{du} = \dfrac{1}{2}u^{-(1/2)} = \dfrac{1}{2u^{(1/2)}} = \dfrac{1}{2\sqrt{u}} = \dfrac{1}{2\sqrt{2x^3 - 4x + 5}}.\)

That gives you your "left hand factor."

But the derivative you want is \(\displaystyle \dfrac{dh}{dx}.\)

The chain rule tells you that \(\displaystyle \dfrac{dh}{dx} = \dfrac{dh}{du} * \dfrac{du}{dx} = \dfrac{1}{2\sqrt{2x^3 - 4x + 5}} * (6x^2 - 4) = \dfrac{3x^2 - 2}{\sqrt{2x^3 - 4x + 5}}.\)

Now you do not have to use a formal substitution. You can see inside your head that the expression inside the square root can be treated according to the power rule as a "unit" and so directly apply the power rule to that "unit" and then, realizing that you need to apply the chain rule to get a derivative in terms of your independent variable, multiply the result of the power rule times the derivative of the "unit" with respect to the independent variable. But if you want to understand what you are doing, you need to realize that you are implcitly doing a substitution in your head.

Furthermore, you will find that the chain rule may need to be used three, four, even five times, in which case your "left-hand" and "right-hand" terminology will become meaningless

 

[JeffM]

Not sure if this constituted a new thread, so put it on this one:
However, regarding another problem:

2.)

\(\displaystyle y = sqrt/(3x^{2} + 4)\)

The book's answer isn't what it seems to be leading to.

Books answer:

\(\displaystyle y = 3x(3x^{2} +4)^{1/2}(3x^{2} + 4)^{-1}\)

Which of course, as before, can be written as a fraction with the negative exponated (if that's the right word) linear factor in the bottom. The linear factor with the exponent of positive 1/2 could be re-written as a radical, as we know.

Please do not add new problems to an old thread, and certainly do not edit an old post to add a new problem.

The way the book has written this answer is odd, but correct. You have not shown what you believe the correct answer should be. Why don't you try working it out using the explicit substitution method I have now shown you twice and see whether you get an answer that is the same or has the same meaning as the book's answer.

 

[mmm4444bot]

\(\displaystyle y = (2x^{3} - 5x^{2} + 4)^{5}\)

turn the 5 exponent (which is outside the parenthesis) into a 5 coefficient (in front of the parenthesis). However, keep an exponent outside the parenthesis and make it 4.

My own words, but new to this stuff. The book is a new current book from the bookstore.

Okay. Maybe you were taught some weird wordings, in the past. The first quote above shows one example of funky description.

We're not "turning exponents into coefficients", when using the Power Rule. We don't "keep an exponent", either. (What if the new exponent were zero? Why "keep" it?)

Here's a better way to describe the Power Rule (less wordy and more accurate):

"Multiply by the exponent, then reduce the exponent by 1."

a^5 becomes 5*a^(5-1) = 5a^4

(In your exercise, a = 2x^3 - 5x^2 + 4)

I hope that your new book is better at describing things than the web site that you were using earlier. :cool:

 

[Jason76]

The problems were from the book, but the wording was from me. But I'm new to this stuff.

 

[mmm4444bot]

Power Rule: Multiply by exponent; reduce exponent by 1

That's just a sample, but learning correct wordings (terminology), or considering them, may help you to remember rules from reading texts and/or to focus your thinking when attempting to use them. :idea: