Help with Derivative: y = ln[x^4+3x+1]

[swears]

Hi, can someone please tell me what I'm doing wrong.

$$y = ln[x^4 +3x +1]$$

$$y' = \frac {1}{x^4 +3x +1} 4x^3 + 3$$

=$$\frac {4x^3 + 3}{x^4 +3x +1}$$

 

[StatusX]

Nothing. Except for you forced me to write this sentence because "Nothing" by itself is too short a message to post, but that's not really your fault.

 

[swears]

Hmm, I was told it was wrong. Maybe because I skipped that middle step when I showed it to someone.

 

[bomba923]

Hi, can someone please tell me what I'm doing wrong.

$$y = ln[x^4 +3x +1]$$

$$y' = \frac {1}{x^4 +3x +1} 4x^3 + 3$$

=$$\frac {4x^3 + 3}{x^4 +3x +1}$$

Well, you might want to include some parentheses on that second line:

$$\begin{gathered} y = \ln \left( {x^4 + 3x + 1} \right) \hfill \\ y' = \frac{1}{{x^4 + 3x + 1}}\left( {4x^3 + 3} \right)\;\;\;\;\;\; \leftarrow parentheses\;{\text{:D}} \hfill \\ y' = \frac{{4x^3 + 3}}{{x^4 + 3x + 1}} \hfill \\ \end{gathered}$$

 

[swears]

Ok, Thanks Guys.

Maybe you can check this for me, Since I don't have the answer.

$$xy^3 + x - y + 21 = 0$$

$$1y^3 + x3y^2 * y' + 1 - y' = 0$$

$$y'[3xy^2 - 1] = -y^3 - 1$$

$$y' = \frac {-y^3 - 1}{3xy^2 - 1}$$

 

[HallsofIvy]

Looks good to me.

 

[bomba923]

Ok, Thanks Guys.

Maybe you can check this for me, Since I don't have the answer.

$$xy^3 + x - y + 21 = 0$$

$$1y^3 + x3y^2 * y' + 1 - y' = 0$$

$$y'[3xy^2 - 1] = -y^3 - 1$$

$$y' = \frac {-y^3 - 1}{3xy^2 - 1}$$

Just as in my post #4, here's another nitpick:

~You can omit some signs in your answer, writing it as:
$$y' = \frac {y^3 + 1}{1-3xy^2}$$

(Looks cleaner, I think)