# Thread: How do I find the derivative of this equation?

1. ## How do I find the derivative of this equation?

How must I find it?

2. ## Re: How do I find the derivative of this equation?

Originally Posted by bossbasslol
How must I find it?
First, there is no equation in your post. Do you see an equal sign? I don't.
Second, why did you post a question about derivatives in the algebra forum?
Have you studied anything about derivatives? Do you know anything about derivatives?

3. ## Re: How do I find the derivative of this equation?

Honest answer. I forgot to input the equal sign, which forms the following equation: A(area)=[insert equation here]
Also, I thought derivatives traced back all the way to Algebra II. I do not know nothing about derivatives except that I require them to finish this equation.
Thus, my knowledge led me to believe that it was appropriate to post derivatives in the Algebra section.

So, can you help me?

4. ## Re: How do I find the derivative of this equation?

Originally Posted by bossbasslol
Honest answer. I forgot to input the equal sign, which forms the following equation: A(area)=[insert equation here]
Also, I thought derivatives traced back all the way to Algebra II. I do not know nothing about derivatives except that I require them to finish this equation.
Thus, my knowledge led me to believe that it was appropriate to post derivatives in the Algebra section.
Derivatives are the prime focus of the first course in The Calculus.
If $y=f(x)=kx^n$ then the simple derivative is $\frac{dy}{dx}=f'(x)=n\cdot kn^{n-1}$

If $f(w)=150w^1-\dfrac{70w^2}{40}$ then the derivative is $150(1)w^{1-1}-\dfrac{2(70)w^{2-1}}{40}=150-\dfrac{70w}{20}$

5. ## Re: How do I find the derivative of this equation?

Thank you, but I forgot to include that the derivative must be for d/dw. I got a different answer, 150 - (7w/2).

6. ## Re: How do I find the derivative of this equation?

Then, frankly, you shouldn't be taking a course that requires you to take derivatives. You have made it clear that you don't know what derivatives are and that you have difficulty with algebra. And if you do not know that $\displaystyle \frac{70}{20}$ is exactly the same $\displaystyle \frac{7}{2}$ then you need to review arithmetic as well.

7. ## Re: How do I find the derivative of this equation?

I apologize for my ignorance on the fractions. But I know most of my algebra, however, the topics of derivatives were not covered in my courses. I'm reviewing them right now .

8. ## Re: How do I find the derivative of this equation?

Originally Posted by bossbasslol
IBut I know most of my algebra, however, the topics of derivatives were not covered in my courses. I'm reviewing them right now .
The topic of derivatives should not be part of a basic algebra course. That begs a question: Why are you reviewing a topic that is not part of your course? In fact it is a topic that is beyond the scope of your studies. There is something that is really suspicious about this thread. Do you want to tell us what is going on?

9. ## Re: How do I find the derivative of this equation?

Yes. I forgot to mention that Algebra isn't the only thing being covered in our course. But it is my terrible mistake to just dismiss derivatives (in which they are more advanced that Algebra) as algebra material.

10. ## Re: How do I find the derivative of this equation?

In all my years I've never encountered an Algebra 2 class that would cover a Calculus topic.

Either way, I have moved the thread to the Calculus forum.

-Dan

11. ## Re: How do I find the derivative of this equation?

Originally Posted by bossbasslol
Honest answer. I forgot to input the equal sign, which forms the following equation: A(area)=[insert equation here]
Also, I thought derivatives traced back all the way to Algebra II. I do not know nothing about derivatives except that I require them to finish this equation.
Thus, my knowledge led me to believe that it was appropriate to post derivatives in the Algebra section.

So, can you help me?
Is it possible you have been asked to find the minimum of the given quadratic function, and you did some searching and found that derivatives can be used for that? There are ways to do this specifically for quadratic functions that don't involve differential calculus.

12. ## Re: How do I find the derivative of this equation?

Yes! Thank you for the clear explanation, I just hate using quadratic to vertex form. I found out that derivative form is much easier, hence my understanding of the need to use it.

13. ## Re: How do I find the derivative of this equation?

Originally Posted by bossbasslol
Yes! Thank you for the clear explanation, I just hate using quadratic to vertex form. I found out that derivative form is much easier, hence my understanding of the need to use it.
Well damn you. Why did you not say that to begin with?
In other words, you are cheating with a shortcut. You think that you know more that the course designer!
Did it ever occur to you that there may be a very good reason why one may want to find a max/min using algebraic methods?
If you are in an algebra course the use algebra methods. You are CHEATING otherwise. If I find your logon, I will tell your instructor.

14. ## Re: How do I find the derivative of this equation?

Originally Posted by bossbasslol
Yes! Thank you for the clear explanation, I just hate using quadratic to vertex form. I found out that derivative form is much easier, hence my understanding of the need to use it.
I should have said maximum instead of minimum. We see that we have a quadratic whose graph is a parabola opening downwards.

I would begin by reducing the fraction:

$\displaystyle A(w)=150w-\frac{7}{4}w^2$

Now, let's factor:

$\displaystyle A(w)=\frac{1}{4}w(600-7w)$

In this form, it's east to see the roots are:

$\displaystyle w\in\left\{0,\frac{600}{7}\right\}$

Now, the axis of symmetry will be midway between the roots, or

$\displaystyle w=\frac{300}{7}$

And so we can state:

$\displaystyle A_{\max}=A\left(\frac{300}{7}\right)= 150\left(\frac{300}{7}\right)- \frac{7}{4}\left(\frac{300}{7}\right)^2= \frac{2}{7}150^2-\frac{150^2}{7}= \frac{150^2}{7}$

As you mentioned, putting the quadratic into vertex form will work

$\displaystyle A(w)=-\frac{7}{4}w^2+150w=-\frac{7}{4}\left(w^2-\frac{600}{7}+\left(\frac{300}{7}\right)^2\right)+ \frac{7}{4}\left(\frac{300}{7}\right)^2=-\frac{7}{4}\left(w-\frac{300}{7}\right)^2+\frac{150^2}{7}$

This tells us:

$\displaystyle A_{\max}=\frac{150^2}{7}$

And also we know that if given:

$\displaystyle y=ax^2+bx+c$

Then the axis of symmetry lies along the line:

$\displaystyle x=-\frac{b}{2a}$

Applying this to $\displaystyle A(w)$ we find the axis of symmetry is:

$\displaystyle w=-\frac{150}{2\left(-\dfrac{7}{4}\right)}=\frac{300}{7}$

And then we proceed as in the first method I showed above.

Personally, I would wait to study derivatives until you are in a calculus course.

15. ## Re: How do I find the derivative of this equation?

Vertex formula is just too tedious a task to accomplish. Which is why I got creative and researched a new way to do it, and my research lead me to follow the derivative method. I must say it is truly amazing. But I just like using d/dx on quadratics.

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