Logarithm differentiation + chain rule

[JamesGoh]

For this function

y=\sqrt{2ln(x)+1}

if I use the chain rule properly, should I be getting this answer?

\frac{dy}{dx}=\frac{2}{x} \times \frac{1}{2} \times \frac{1}{\sqrt{2ln(x)+1}}

My aim of doing this is to verify that

\frac{dy}{dx}=\frac{1}{xy}

 

[ShayanJ]

That's right and equal to what you want!

 

[SteliosVas]

That's right and equal to what you want!

Exactly what Shaun said.
If you equate that to what information you need.

BINGO! You will get your answer

 

[Mark44]

As a simpler alternative to using the chain rule, you can do this:

$y = \sqrt{2\ln(x) + 1}$
$\Rightarrow y^2 = 2\ln(x) + 1$
Differentiating implicitly,
$2y\frac{dy}{dx} = \frac{2}{x}$
$\Rightarrow \frac{dy}{dx} = \frac{1}{xy}$

 

[Apogee]

Yes, that is correct. Just to show the math:

d/dx[√2ln(x)+1]
=1/2√2ln(x)+1 * 2/x
=2/x * 1/2 * 1/√2ln(x)+1

 

[Mark44]

You are missing several sets of parentheses.

Yes, that is correct. Just to show the math:

d/dx[√2ln(x)+1]

Most would interpret the part in brackets above as √2 * ln(x)+1, rather than √(2ln(x)+1).

=1/2√2ln(x)+1 * 2/x

Most would interpret the above as
1/2 * √2ln(x) + 2/x, which I don't think is what you intended.

=2/x * 1/2 * 1/√2ln(x)+1

 

[Apogee]

You're right, perhaps I should have put 2ln(x) + 1 in parentheses, but I think he gets what I mean haha. Nonetheless, I appreciate the correction. Thank you!

 

[paradoxymoron]

As a simpler alternative to using the chain rule, you can do this:

$y = \sqrt{2\ln(x) + 1}$
$\Rightarrow y^2 = 2\ln(x) + 1$
Differentiating implicitly,
$2y\frac{dy}{dx} = \frac{2}{x}$
$\Rightarrow \frac{dy}{dx} = \frac{1}{xy}$

Hm, how does this take into account the fact that squaring the equation will give multiple values?

$y^2=2\ln(x)+1$ is a set of two equations, namely $$y= \sqrt{2\ln(x) + 1}$$ and $$y= -\sqrt{2\ln(x) + 1}$$
Will the derivative also be multiple valued? It doesn't seem like it since $\frac{1}{xy}$ is a singled valued funtion. What am I missing?

 

[Mark44]

As a simpler alternative to using the chain rule, you can do this:

$y = \sqrt{2\ln(x) + 1}$
$\Rightarrow y^2 = 2\ln(x) + 1$
Differentiating implicitly,
$2y\frac{dy}{dx} = \frac{2}{x}$
$\Rightarrow \frac{dy}{dx} = \frac{1}{xy}$

Hm, how does this take into account the fact that squaring the equation will give multiple values?

Actually it doesn't in this case. That's something that I considered but didn't mention before. In the first equation, the square root term is necessarily nonnegative, which means that y must also be nonnegative. So squaring both sides doesn't introduce extraneous solutions that weren't present in the original equation.

$y^2=2\ln(x)+1$ is a set of two equations, namely $$y= \sqrt{2\ln(x) + 1}$$ and $$y= -\sqrt{2\ln(x) + 1}$$
Will the derivative also be multiple valued?

I don't understand what you're asking. The derivative will not be multiple valued.

It doesn't seem like it since $\frac{1}{xy}$ is a singled valued funtion. What am I missing?

 

[paradoxymoron]

I don't understand what you're asking. The derivative will not be multiple valued.

What I'm trying to ask is that since we found the derivative is $\frac{dy}{dx}=\frac{1}{xy}$, we have two options for $y$ now, the positive or negative root, since we squared the term to make the differentiation simpler. Why do we necessarily choose the positive one? I know I'm just being technical, sorry about my arrogance, but it's a pet-peeve of mine.

 

[Mark44]

What I'm trying to ask is that since we found the derivative is $\frac{dy}{dx}=\frac{1}{xy}$, we have two options for $y$ now, the positive or negative root, since we squared the term to make the differentiation simpler.

No we don't in this case, which I explained in post #9. Since y is the square root of some expression, y is necessarily nonnegative. Take a look again at what I wrote, and see if that answers your questions.

Why do we necessarily choose the positive one? I know I'm just being technical, sorry about my arrogance, but it's a pet-peeve of mine.