1. Feb 28, 2013

### jkristia

I'm preparing for a test next week and am working on some practice problems we got, but we did not get the answers. There are 2 problems where I'm not sure if I have done it correct or not and would like to have it checked.

This is the first problem. I think I have followed the chain, product and exponential rules, but I really have no idea if this is correct or not.

Any help is appreciated.

#### Attached Files:

• ###### derivative_1.png
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2. Feb 28, 2013

### Mute

Yep, looks correct.

3. Feb 28, 2013

### jkristia

wow, thank you very much for the quick reply

4. Mar 1, 2013

### jkristia

Instead of starting a new thread, I will post the second problem here.
I think I have done this one correct, but as with the first problem, I really have no idea if it is correct or not

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5. Mar 1, 2013

### haruspex

That's correct, but you could go further and use the second line of that to substitute for y in the last line, so you have y' entirely in terms of x. Don't know if that's what was wanted.

6. Mar 1, 2013

### Staff: Mentor

In the future, when you post a new problem, please start a new thread.

7. Mar 1, 2013

### jkristia

>>In the future, when you post a new problem, please start a new thread
ok, I will remember that

>>That's correct, but you could go further and
Good point. But in this case since I don't start out with a y = ... , but rather yx = ..., then I'm not sure if it would be expected.
If I were to replace it, I should replace y for e$\frac{tan(x) ln(x)}{x}$, correct?.
Anyway, I will try and keep this in mind for the test.
Thank you very much for your help

8. Mar 1, 2013

### haruspex

Yes, or $y = x^{\frac{tan(x)}{x}}$. You could reduce it to $y' = x^{\frac{tan(x)}{x}-1}[\frac{tan(x)}{x}\left(1-ln(x)\right)+ln(x)sec^2(x)]$