Derivative using the Quotient Rule

[Sap67]

Greetings, I've been having some trouble with this one function where I need to get its derivative, I tried using a derivative calculator but my answer never matches up with what the calculator gets. Plus, the calculator never simplifies it until the end so I don't have any idea if it's right or wrong. The step by step instructions aren't really helping either since they're using some things I don't fully comprehend.
I'm using the quotient rule, I think that's its name, I'm referring to the following formula : (u'v.v'u)/v².
I used this formula : u'/2√u for the square root part in the denominator.

 

[tkhunny]

Have you considered the equivalent derivative of [math]\dfrac{4x^{5}-23x^{4}+40x^{3}-20x^{2}-4x+3}{x^{2}-4x+3}[/math]?

You should already have noticed that the derivative fails to exist on [1,3].

I'm guessing that's a "no", but I thought I'd mention it just to emphasize that this is an AMAZXINGLY tedious problem. Are you SURE you're in the right problem set? Yikes.

 

[skeeter]

numerator and denominator have equal roots [MATH]x \in \{1,3 \}[/MATH]
quotient equals [MATH]4x^3-7x^2+1 \, , \, x \cancel{\in} \{1,3 \}[/MATH]

 

[pka]

Greetings, I've been having some trouble with this one function where I need to get its derivative, I tried using a derivative calculator but my answer never matches up with what the calculator gets. Plus, the calculator never simplifies it until the end so I don't have any idea if it's right or wrong. The step by step instructions aren't really helping either since they're using some things I don't fully comprehend.
I'm using the quotient rule, I think that's its name, I'm referring to the following formula : (u'v.v'u)/v².
I used this formula : u'/2√u for the square root part in the denominator.

Here is a PDF file of a complete solution by WolframAlpha.

 

[skeeter]

didn’t catch the radical in the OP ... oops.

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[Sap67]

Yes, the problem is right, here's the sheet with the questions, it's in french but I'm posting it to prove that this is what I need to do.
First of all, before answering to all of you, I need to say, there's a lot of what you're saying that I don't understand, things I simply didn't learn. I'm a very serious student so it's not by lack of studying that I don't know these things, it's because we simply have not learned them, for whatever reason. So I'll begin

You should already have noticed that the derivative fails to exist on [1,3].

I'm sorry, but what does exist on 1,3 mean ? are 1 and 3 the x and y for the slope point ? If so I didn't get them as the exercice didn't ask for those, it only asked to get the derivative.

Here is a PDF file of a complete solution by WolframAlpha.

I really appreciate it, thank you but I've tried relying on the calculators but they use techniques I haven't learned and I probably can't use as my teacher would probably accuse me of just copying of the internet, even if I do learn how those techniques work. Like I have no clue what that "d/dx" symbol means as we've just never used it before. Nor have we learned about the chain rule, and we don't use powers that are like fractions like 3/2.
I probably sound dumb, but we just never learned all of those, probably because I'm in a technology course so our Math is simplified compared to the rest.

 

[pka]

I've tried relying on the calculators but they use techniques I haven't learned and I probably can't use as my teacher would probably accuse me of just copying of the internet, even if I do learn how those techniques work. Like I have no clue what that "d/dx" symbol means as we've just never used it before. Nor have we learned about the chain rule, and we don't use powers that are like fractions like 3/2.

You do not sound dumb at all. Given what you posted, one must infer that the course is poorly conceived in the extreme. Now you must take action by going to the head of your school and filling a complaint. If you don't then you might be thought of as a bit dumb.

 

[skeeter]

[MATH]y=\dfrac{4x^3-7x^2+1}{\sqrt{x^2-4x+3}}[/MATH]
Note the denominator equals [MATH]\sqrt{(x-3)(x-1)}[/MATH] which means the function is undefined for all values in the interval [MATH]1 \le x \le 3[/MATH] because the endpoints make the denominator equal zero, and all the values of x between those endpoints make the value of the expression under the radical negative. This is recognition of a function’s domain, a topic that should be covered before calculus.

[MATH]y’ = \dfrac{\sqrt{x^2-4x+3} (12x^2-14x) - (4x^3-7x^2+1) \frac{x-2}{\sqrt{x^2-4x+3}}}{x^2-4x+3}[/MATH]
multiply numerator and denominator by [MATH]\sqrt{x^2-4x+3}[/MATH] ...

[MATH]y’ = \dfrac{(x^2-4x+3)(12x^2-14x) - (4x^2-7x^2+2)(x-2)}{(x^2-4x+3)^{3/2}}[/MATH]
from here, you can multiply out the numerator and combine like terms.

 

[pka]

didn’t catch the radical in the OP ... oops.

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I think that this is a perfect example of why this forum should ban positing problems(questions) in images.
These images are almost always posted in the wrong direction for reading.
Require questions to be posted in either normal type-face or better in LaTeX coding.