Calculus 1 - 3 snags (work included), could someone double check me on these?

Oct 2015
1
0
Idaho
Hi everyone! I'm taking a Calculus course and have hit a couple snags. The following are the problems I have (first two are to be implicitly differentiated, the third differentiated normally by use of the quotient rule), followed by what I tried to solve each. If anyone could guide me so that I understand this a little better, you'd be awesome! :D


(1 + e^(3x))^2 = 3 + ln(x + y)

and

e^(x - y) = ln(x - y)

and

y = (x^4 + 3x^2 + x)/(ln x)

For the first one, I've tried 2(1 + e^(3x))^1 = 0 + 1/(x + y). I'm not so sure I'm going about this the correct way.

For the second one, I got e^(x - y) = 1/(x - y).

And for the third, I understand you have to do the quotient rule. By that I got the setup. But before continuing, would it be correct to make the denominator x^-1 for when I bring it to the top to carry out it's derivative against the top? I've asked a friend how they did this problem, they said they are going to ask the teacher. I'd like to understand in advance if possible.

I really do appreciate the help, guys. Thanks again. :)
 
Sep 2015
11
2
Seoul
Make sure you don't forget to differentiate \(\displaystyle y\) by \(\displaystyle x\) to get \(\displaystyle \frac{dy}{dx}\) as \(\displaystyle y\) is not a constant, but a function of \(\displaystyle x\).

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#1.

\(\displaystyle (1+e^{3x})^{2}=3+\textup{ln}(x+y)\)

\(\displaystyle 2(1+e^{3x}) 3e^{3x} = \frac{1+\frac{dy}{dx}}{x+y}\)

\(\displaystyle \therefore \frac{dy}{dx} = 6e^{3x}(1+e^{3x})(x+y)-1\)

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#2.

\(\displaystyle e^{x-y}=\textup{ln}(x-y)\)

\(\displaystyle \left(1-\frac{dy}{dx}\right)e^{x-y}=\frac{1-\frac{dy}{dx}}{x-y}\)

\(\displaystyle \therefore \frac{dy}{dx} = 1\)

(I'm not too sure about this.)
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#3.

\(\displaystyle y=\frac{x^4+3x^2+x}{\textup{ln}x}\)

\(\displaystyle y'=\frac{(4x^3+6x+1)(\textup{ln}x)-(x^4+3x^2+x)\left(\frac{1}{x}\right)}{(\textup{ln}x)^2}\)

\(\displaystyle y'=\frac{4x^3+6x+1}{\textup{ln}x}-\frac{x^3+3x+1}{(\textup{ln}x)^2}\)
 
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