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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 = .....

janhardo
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1 Answers1

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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]))

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

Syed
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  • 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 – janhardo Apr 30 '22 at 14:13
  • This will work for separable equations only. I chose 0 as I wanted to Solve and it needs an equation. – Syed Apr 30 '22 at 14:25
  • Ah , yes the differential of a number (contant) is 0. A differential equation with variables to be separated is this type. – janhardo Apr 30 '22 at 14:39