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The idea is here to have a implicit equation and take from both sides of the equation the differential and solve out of this the derivative

The derivative is the ratio of two differentials and can be thought of as a fraction : dy/dx

eqn = Dt[x*y^3 + Sin[x + y]] == Dt[2]

note: (dx+dy) = d(x+y) ( a calculation rules of differentials (one of them))

Using Solve for getting dy and dy is not working ?

The desired expression must be dy/dx = .....

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    $\begingroup$ An equation has 2 sides, you only specify one. $\endgroup$
    – Daniel Huber
    Apr 30 at 12:19
  • $\begingroup$ Thanks, yes i did first for Solve[ eqn, Dt[x]] , but Solve gives a error $\endgroup$
    – janhardo
    Apr 30 at 12:36
  • $\begingroup$ I do see now what you mean: i forget the RHS $\endgroup$
    – janhardo
    Apr 30 at 14:48
  • $\begingroup$ I added it later the RHS in post .. you are right , thanks $\endgroup$
    – janhardo
    Apr 30 at 15:08
  • $\begingroup$ You're welcome. $\endgroup$
    – Michael E2
    Apr 30 at 15:10

1 Answer 1

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expr = y^3 Dt[x] + 3 x y^2 Dt[y] + Cos[x + y] (Dt[x] + Dt[y])

(Dt[y] /. First@Solve[expr == 0, Dt[y]])/Dt[x] // Simplify

-((y^3 + Cos[x + y])/(3 x y^2 + Cos[x + y]))

$$-\frac{\cos (x+y)+y^3}{3 x y^2+\cos (x+y)}$$

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  • $\begingroup$ Syed thanks great. On this code I would not come up with myself. Its not clear why this expr must be zero ( homogen? ) But, wait i see i forget something in my example! In my example to get the derivative from an implicit equation ( two methods) and this is via by taken the differentials from both sides of the equation. Its not 0, but it is 2 in the differential equation $\endgroup$
    – janhardo
    Apr 30 at 14:13
  • $\begingroup$ This will work for separable equations only. I chose 0 as I wanted to Solve and it needs an equation. $\endgroup$
    – Syed
    Apr 30 at 14:25
  • $\begingroup$ Ah , yes the differential of a number (contant) is 0. A differential equation with variables to be separated is this type. $\endgroup$
    – janhardo
    Apr 30 at 14:39

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