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Can you do this derivative????


Mig-El

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Hi, well I'm not so sure about other countries but around here math is very much hated. That's not my case. The thing is that my teacher apllied us this exam with stuff we didn't even practice at all, he just mentioned the formulas and explained the tiniest example of each, now (after the exam) he's started with it and he just gave us an exercise and told me to do it in 2 ways, with the formula and with logarithm propertys, can you tell me how to do it with the second cause the way I did was wrong according to him.

 

y= x√x

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What the hell are you talking about>? Im good at maths but I'm probably not advanced enought to understand what you are talking about.

Please don't get me wrong, what I meant is that back here like 2 off 40 studentes like math hahah, I don't know in other places but around here people don't like math that much, I do like math and I'm good actually but I'm having a little problem here

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Haha I am probably the LAST person he'd want help from though...I suck at math. :P But I'll try.

 

Mig-El, do you happen to have the answer of the problem? To make sure that I'm actually right? Haha I have solved this problem numerous times in class but I can't ever remember the exact answer. :lol:

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derivative.jpg

 

*exhale* okay, here it goes:

 

So you start out with the original problem. You want the derivative, right?

 

1. Another way to write the square root of x is x to the 1/2 power. This will make getting the derivative easier.

 

2. Since the x is now in parentheses when I put it in the 1/2 power, it is an indication of using the chain rule. You probably already know what that is, but you multiply the 1/2 with the first x, and subtract 1 from the exponent. So now it's 1/2x (x) to the -1/2 power. Along with that, you have to multiply the whole thing with the derivative of the x in the parentheses, which is 1, so nothing changes.

 

3. I rewrote it so that the answer would be a fraction...1/2x turns into x/2....and since anything with a negative exponent goes into the denominator....it turns back to the square root, but in the bottom. So the answer would be x/2 square root x....

 

...I think. :lol: I hope that makes sense and I hope it's right. You'll have to tell me haha.

 

Oh, and sometimes teachers don't like it when the radicals are in the denominator (I don't care, it's more work for me...:P) But at the bottom I simplified it more so that there's no radical at the bottom...but if that's not important then don't worry about it.

 

Geez, I suck at math, haha. But I hope I was right!

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derivative.jpg

 

*exhale* okay, here it goes:

 

So you start out with the original problem. You want the derivative, right?

 

1. Another way to write the square root of x is x to the 1/2 power. This will make getting the derivative easier.

 

2. Since the x is now in parentheses when I put it in the 1/2 power, it is an indication of using the chain rule. You probably already know what that is, but you multiply the 1/2 with the first x, and subtract 1 from the exponent. So now it's 1/2x (x) to the -1/2 power. Along with that, you have to multiply the whole thing with the derivative of the x in the parentheses, which is 1, so nothing changes.

 

3. I rewrote it so that the answer would be a fraction...1/2x turns into x/2....and since anything with a negative exponent goes into the denominator....it turns back to the square root, but in the bottom. So the answer would be x/2 square root x....

 

...I think. :lol: I hope that makes sense and I hope it's right. You'll have to tell me haha.

 

Oh, and sometimes teachers don't like it when the radicals are in the denominator (I don't care, it's more work for me...:P) But at the bottom I simplified it more so that there's no radical at the bottom...but if that's not important then don't worry about it.

 

Geez, I suck at math, haha. But I hope I was right!

 

You lost me

 

 

I just got lost!...

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Are you in Calculus, Mig-El? I don't remember using logarithms to find derivatives...unless there is such thing as taking a derivative of a logarithm...which I haven't done yet so whoops. :P

 

Haha thanks, Lore, but I don't think I helped. :lol: I might be talking about a completely different thing.

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