Question

Can someone help me with implicit differentiation please:

(xy)^1/2 + 2x = y^1/2

Answer 1

using the cahin rule, and the Leibniz rule, we can write this:
\[\Large \frac{{d\sqrt {xy} }}{{dx}} = \left( {\sqrt {xy} } \right)' = \frac{1}{2}\frac{1}{{\sqrt {xy} }}\left( {y + xy'} \right)\]

Answer 2

oops..chain*

Answer 3

being y=y(x), namely y is a function of x

Answer 4

I am a bit lost on how you get (y+xy')

Answer 5

since we hav to compute the first derivative of this function:
\[\Large {xy}\]

Answer 6

using the product rule or Leibniz rule

Answer 7

have*

Answer 8

ahhhhh I got it thank you :)

Answer 9

thank you! :)

Answer 10

then is it + 2

Answer 11

if we differentiate all of your terms, we get:
\[\Large \frac{1}{2}\frac{1}{{\sqrt {xy} }}\left( {y + xy'} \right) + 2 = \frac{1}{2}\frac{1}{{\sqrt y }}y'\]

Answer 12

now, the least common multiple is:
\[\Large \sqrt {xy} = \sqrt x \sqrt y \]

Answer 13

so we have to multiply both sides of that expression, by :
\[\Large\sqrt x \sqrt y \]

Answer 14

what do you get?

Answer 15

1/2(y+xy')+2=x^2sqrty/2 y'
??

Answer 16

I got this:
\[\Large \frac{1}{2}\left( {y + xy'} \right) + 2\sqrt {xy} = \frac{1}{2}\sqrt x y'\]

Answer 17

yes ok I understand that :)

Answer 18

really, the least common multiple was
\[\Large 2\sqrt {xy} = 2\sqrt x \sqrt y \]
nevertheless never mind, since we can multiply, both sides of last expression, by 2 right now

Answer 19

so, if we multiply both sides of my last expression by 2, we get:
\[\Large y + xy' + 4\sqrt {xy} = \sqrt x y'\]

Answer 20

yes I got that as well :)

Answer 21

now we subtract, from both sides this quantity:
\[\Large xy'\]
so we can write:
\[\Large y + 4\sqrt {xy} = \sqrt x y' - xy'\]

Answer 22

then take out the y' as the common factor?

Answer 23

better is to factor out:
\[\Large \sqrt x y'\]
at the right side

Answer 24

so we can write:
\[\Large y + 4\sqrt {xy} = \sqrt x y'\left( {1 - \sqrt x } \right)\]

Answer 25

then divide the other side by sort x(1-sqrtx)?

Answer 26

oops not sort sqrt

Answer 27

yes! we have to divide both sides, by:
\[\Large \sqrt x \left( {1 - \sqrt x } \right)\]

Answer 28

so we can write the expression for y', what do you get?

Answer 29

\[y'=\frac{ y+4\sqrt{xy} }{ (\sqrt{x} -x)}\]

Answer 30

that's right!

Answer 31

Are you able to walk me through a couple more if you have time, I am still struggling to get my head around them :)

Answer 32

ok!

Answer 33

Thank you so very much I really appreciate it, I can close this and put up a new one so I can give you another medal :)

Answer 34

ok! :)