# Logarithm differentiation + chain rule

## Main Question or Discussion Point

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
Gold Member
That's right and equal to what you want!

That's right and equal to what you want!

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

Mark44
Mentor
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,
$2y\frac{dy}{dx} = \frac{2}{x}$
$\Rightarrow \frac{dy}{dx} = \frac{1}{xy}$

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
Mentor
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

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!

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 in to 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
Mentor
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 in to 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?

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
Mentor
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.