## marigirl one year ago Implicit differentiation: I understand that you would need to use it for cases where the question is to differentiate y=x^x But how do i know when to use it? Could you see the picture i attached below, implicit differentiation was used, however, i could have made it the subject and gone about differentiating it explicitly..

1. marigirl

When i did this question, i made r the subject by taking the square root of the left side

2. zepdrix

When to use it? Hmm

3. zepdrix

You use it when isolating y is more work than it's worth, sometimes it's even impossible

4. zepdrix

But for your problem, you're saying that end up with a square root which seems different from the correct answer or something?

5. marigirl

okay, and yes in my question What i did was i did $r=\sqrt{60h-h^2}$ and i thought that was not too bad to differentiate......but the model answers went about implicit differentiating it

6. zepdrix

Well you get something that is not a function when you solve for r directly,$\large r=\pm\sqrt{60h-h^2}$Don't forget the +/- when you take the root of a square! :) Maybe that's why they avoided going that way, hmm.

7. zepdrix

So doing it implicitly, you end up with:$\large\rm r'=\frac{30-h}{r}$If we plug r into this equation:$\large\rm r'=\pm\frac{30-h}{\sqrt{30^2-(30-h)^2}}$It probably looks more or less the same as what you got, going the other route, yes? :o

8. marigirl

yes ..so i worried about nothing

9. zepdrix

XD

10. marigirl

Thanks :) I appreciate it

11. zepdrix

This MIT video gives an example of this type of thing happening https://youtu.be/5q_3FDOkVRQ?t=9m10s Just in case you wanted to have a better understanding of what is going on, he can explain it much better than I can :D lol

12. marigirl

I wish to be confident like you in maths one day :) your truly an inspiration

13. zepdrix

aw :3 lol

14. marigirl

hey

15. marigirl

$y=x^{\frac{ m }{ n }}$ how did that become $y^n=x^m$

16. zepdrix

Start with a simple example to convince yourself:$\large\rm y=\sqrt{x}$If I rewrite this as:$\large\rm y=x^{1/2}$In either form, I can see that I need to square to get rid of the root,$\large\rm y^2=x$This will happen with any degree of root, your exponent is always the reciprocal of the power on x.$\large\rm y=\sqrt[3]{x}\qquad\to\qquad y=x^{1/3}\qquad\to\qquad y^3=x$Err I guess I should be careful the way I say reciprocal, because we really don't care about the numerator on x. So back to the example you gave:$\large\rm y=x^{m/n}$To deal with this n'th root, I'm raising both sides to the n'th power,$\large\rm y^n=\left(x^{m/n}\right)^n$Rules of exponents gives:$\large\rm y^n=x^{\frac{m}{n}\cdot n}$$\large\rm y^n=x^m$

17. marigirl

yes i see it.. thank you so much. im off to watch more MIT videos now :D

18. zepdrix

c: