Derivatve of function

[Anakin_99]

Hi guys I can't figure out what is wrong in this derivate, my prof solves it by collecting (3x - 1) at nominator and denominator but I'd like to understand what is wrong in my method. Thank u for help and sorry if my english i so bad. Result should be
(6x+1) /
(3x(3x−1))

 

[lex]

Just your very last fraction!
[MATH]\frac{-2-\frac{1}{3x}}{-1+3x}[/MATH]
Multiply top and bottom by [MATH]3x[/MATH]
[MATH]\frac{-(6x+1)}{3x(3x-1)}[/MATH]

 

[Anakin_99]

Omg thank u man in just too tired shouldn t do math for so many hours ahahah how bad i am ahahah

 

[lex]

Sounds like time for a break then!

 

[Steven G]

That last line was not so good!

The real question is why are you using the quotient rule at all??

$$y= \ln(\dfrac{x^{1/3}}{3x-1}) = \ln(x^{1/3}) - \ln(3x-1) = \dfrac{1}{3}\ln(x) - \ln(3x-1)$$
$$y'= \dfrac{1}{3x} - \dfrac{1}{3x-1}$$ * 3 ........................................................(edited)

Did your teacher really solve this by using the quotient rule?? Why??

 

[JeffM]

[MATH]y = ln \left ( \dfrac{x^{1/3}}{3x - 1} \right ).[/MATH]
Now I would do it this way.

[MATH]u = \dfrac{x^{1/3}}{3x - 1} \implies y = ln(u), \ \dfrac{dy}{du} = \dfrac{1}{u} = \dfrac{3x - 1}{x^{1/3}}, \text { and } u^3 = \dfrac{x}{(3x - 1)^3}.[/MATH]
Now I do not have to mess around with fractional exponents. Furthermore

[MATH]\dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} [/MATH] by the chain rule.

[MATH]u^3 = \dfrac{x}{(3x - 1)^3} \implies 3u^2 * \dfrac{du}{dx}= \dfrac{1(3x - 1)^3 - x(3)(3x - 1)^2(3)}{(3x - 1)^6} = \dfrac{(3x - 1)^2\{1(3x - 1) - 9x\}}{(3x - 1)^6} = - \dfrac{6x + 1}{(3x - 1)^4} \implies [/MATH]
[MATH]3 \left ( \dfrac{x^{1/3}}{3x - 1} \right )^2 \dfrac{du}{dx} = - \dfrac{6x + 1}{(3x - 1)^4} \implies \dfrac{3x^{2/3}}{(3x - 1)^2} * \dfrac{du}{dx} = - \dfrac{6x + 1}{(3x - 1)^4} \implies \dfrac{du}{dx} = - \dfrac{6x + 1}{3x^{2/3}(3x - 1)^2}. [/MATH]
[MATH]\therefore \dfrac{dy}{dx} = \dfrac{3x - 1}{x^{1/3}} * \left (- \dfrac{6x + 1}{3x^{2/3}(3x - 1)^2} \right ) = - \dfrac{6x + 1}{3x(3x - 1)}. [/MATH]
Break a problem up to avoid errors in algebra. I guarantee if I did it in one fell swoop, I’d screw it up.

 

[lex]

$$y'= \dfrac{1}{3x} - \dfrac{1}{3x-1}$$

$$y'= \dfrac{1}{3x} - \dfrac{\boldsymbol{3}}{3x-1}$$
Yes. A much neater method.

 

[JeffM]

That last line was not so good!

The real question is why are you using the quotient rule at all??

$$y= \ln(\dfrac{x^{1/3}}{3x-1}) = \ln(x^{1/3}) - \ln(3x-1) = \dfrac{1}{3}\ln(x) - \ln(3x-1)$$
$$y'= \dfrac{1}{3x} - \dfrac{1}{3x-1}$$
Did your teacher really solve this by using the quotient rule?? Why??

Wow. Lovely. I got distracted by looking at what the OP did.

But I think you forgot something.

[MATH]y = ln \left (\dfrac{x^{1/3}}{3x - 1} \right ) = ln(x^{1/3}) - ln(3x + 1) = \dfrac{1}{3} * ln(x) - ln(3x - 1).[/MATH]
[MATH]\therefore y’ = \dfrac{1}{3} * \dfrac{1}{x} - \dfrac{1}{3x - 1} * 3 = \dfrac{3x - 1}{3x((3x - 1)} - \dfrac{9x}{3x(3 - x)} = - \dfrac{6x + 1}{3x(3x - 1)}.[/MATH]
Let’s not tell Subhotosh.

 

[Steven G]

Whenever you have ln(f(x)) you should never use the product rule, quotient rule or power rule. Very powerful.

Now if you want, you NEVER have to use the product rule, quotient rule or power rule again. If you have y = f(x), then simply compute the derivative of ln(y) = ln(f(x))

 

[lookagain]

$$y'= \dfrac{1}{3x} - \dfrac{\boldsymbol{3}}{3x-1}$$
Yes. A much neater method.

When worked this way, this will typically be taken as the final form. As a
grader, I would see that the student was able to know to (at least as an
option) to break apart the logarithmic expression so as to aid in working
the derivative. Then, you don't want to undo the breaking apart by rejoining.
The reward in getting the derivative done by use of the logarithm rules option
should be to keep the algebraic fractions separate/fewer additional steps/less
likelihood of further error.

 

[lex]

When worked this way, this will typically be taken as the final form. As a
grader, I would see that the student was able to know to (at least as an
option) to break apart the logarithmic expression so as to aid in working
the derivative. Then, you don't want to undo the breaking apart by rejoining.
The reward in getting the derivative done by use of the logarithm rules option
should be to keep the algebraic fractions separate/fewer additional steps/less
likelihood of further error.

You seem to be inferring something from my post, that wasn't there.
I simply highlighted that the emboldened '3' was omitted from the numerator of the second expression.

 

[lookagain]

You seem to be inferring something from my post, that wasn't there.
I simply highlighted ...

I know you highlighted.
I used your post, because your answer form is the only one that is correct with the logarithmic method that does
not attempt to join the fractional expressions at the end together as in post # 8, which I am stating defeats the purpose of using the logarithmic method.

 

[lex]

^{I know you highlighted.
I used your post, because your answer form is the only one that is correct with the logarithmic method that does
not attempt to join the fractional expressions at the end together as in post # 8, which I am stating defeats the purpose of using the logarithmic method.}

It seems I was inferring something from your post, that wasn't there!
(I did wonder how you could have come to take that from my post).

 

[JeffM]

@lookagain

No, it is not the only correct answer. 7 + 10 is not the only correct way to give an answer of 17. In fact, the general preference is for simplified expressions. In any case, a result is either correct or not; using a specific method does not make correct conclusions incorrect. The “purpose of the logarithmic method” is to ease derivation, not to prohibit normal simplification of conclusions.

Moreover, there was another purpose in my second post. You cannot get to the correct answer, correct as previously shown, from jomo’s result.

 

[lookagain]

@lookagain

No, it is not the only correct answer. 7 + 10 is not the only correct way to give an answer of 17. In fact, the general preference is for simplified expressions. In any case, a result is either correct or not; using a specific method does not make correct conclusions incorrect. The “purpose of the logarithmic method” is to ease derivation, not to prohibit normal simplification of conclusions.

Moreover, there was another purpose in my second post. You cannot get to the correct answer, correct as previously shown, from jomo’s result.

I did not state it was the only correct answer period. You need to read the rest
of that sentence. Adding the fractions did not simplify the answer. It needlessly
made more work ending in a more complex expression built of more characters
than the two fractions taken together.

Your other purpose does not make sense, because there is no "the correct answer."
There is a correct answer form given by the OP. Remember your first two sentences to me here?

 

[JeffM]

I did not state it was the only correct answer period. You need to read the rest
of that sentence. Adding the fractions did not simplify the answer. It needlessly
made more work ending in a more complex expression built of more characters
than the two fractions taken together.

Your other purpose does not make sense, because there is no "the correct answer."
There is a correct answer form given by the OP. Remember your first two sentences to me here?

Most people consider a single fraction easier to work with than two fractions that need to be added.

But in any case, you said that my answer was not the only correct answer given the method used. That is simply wrong.

Finally, the OP did not give the correct result. Neither did jomo as you would see if you worked out his answer.

 

[Cubist]

Now if you want, you NEVER have to use the product rule, quotient rule or power rule again. If you have y = f(x), then simply compute the derivative of ln(y) = ln(f(x))

Very interesting! It seems to work even if f(x) goes negative. I guess you could prove this by piecewise consideration...

• ln(y) = ln(f(x)) for x∈ {domain where f(x)>0 }
• ln(-y) = ln(-f(x)) for x∈ {domain where f(x)<0 }

...and both give the same result for dy/dx, and so the pieces can be rejoined ( glossing over the fact that x=0 is possible )

 

[lookagain]

77th

Most people consider a single fraction easier to work with than two fractions that need to be added.

The point is there is no need to add the two (or more fractions in other cases)
fractions, because the derivative is finished. Going further adds extra time
that takes away from other problem solving and invites more chances for
errors.

But in any case, you said that my answer was not the only correct answer given the method used. That is simply wrong.

I stated that there was "a correct answer form given by the OP."

Finally, the OP did not give the correct result. Neither did jomo as you would see if you worked out his answer.

Of course I know the OP did not give the correct result. And I know about jomo, too,
because I commented on the corrected result of lex's post of jomo's.

 

[Steven G]

Finally, the OP did not give the correct result. Neither did jomo as you would see if you worked out his answer.

Do you have to rubbing it in? . Only kidding.

 

[lex]

Do you have to rubbing it in? . Only kidding.

It's the (insightful) thought that counts!