Check my implicit differentiation of ln xy + 5x = 30, etc

[warwick]

1) Find dy/dx:

ln xy + 5x = 30

y' = -5y - (y/x)

2) Find dy/dx:

e^(xy) + x^2 - y^2 = 10

I'm in a hurry, so I can't post the work for this one. Thanks for any help.

 

[Unco]

You've done great with the first one, Warwick, so let's see what you've tried for the second. Type out your derivatives term by term and try to show where you get stuck.

 

[warwick]

You've done great with the first one, Warwick, so let's see what you've tried for the second. Type out your derivatives term by term and try to show where you get stuck.

Thanks. Math is not hard for me, but if I don't repeat the work, it tends to slip my mind! Repetition is key!

e^(xy) + x^2 - y^2 = 10 u = xy

e^u du/dx +2x - 2yy' - 0

2yy' = 2x + e^u (xy' +y)

 

[Unco]

e^(xy) + x^2 - y^2 = 10      u = xy

e^u du/dx +2x - 2yy' - 0

2yy' = 2x + e^u (xy' +y)

Excellent work, Warwick. Now all that is left to do is solve for y'.

Group the y' terms to one side:
\(\displaystyle \mbox{2yy' - e^{xy}xy' = 2x + e^{xy}y}\)

Factorise out y':
\(\displaystyle \mbox{y'(2y - e^{xy}x) = 2x + e^{xy}y}\)

And solve for y' from there.