# Logarithm differentiation + chain rule

1. Aug 26, 2014

### 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

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

2. Aug 27, 2014

### ShayanJ

That's right and equal to what you want!

3. Aug 31, 2014

### SteliosVas

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

4. Aug 31, 2014

### Staff: 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,
$2y\frac{dy}{dx} = \frac{2}{x}$
$\Rightarrow \frac{dy}{dx} = \frac{1}{xy}$

5. Sep 1, 2014

### 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

6. Sep 1, 2014

### Staff: Mentor

You are missing several sets of parentheses.
Most would interpret the part in brackets above as √2 * ln(x)+1, rather than √(2ln(x)+1).
Most would interpret the above as
1/2 * √2ln(x) + 2/x, which I don't think is what you intended.

7. Sep 1, 2014

### 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!

8. Sep 2, 2014

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?

9. Sep 3, 2014

### Staff: Mentor

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

10. Sep 4, 2014

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.

11. Sep 5, 2014

### Staff: Mentor

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.