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Here's a homework problem from the Coursera course Calculus: Single Variable by Robert Ghrist

Find the derivative $\frac{dy}{dx}$ from the equation $x \tan y - y^2 \ln x=4$

I wanted to check my work using Mathematica so I tried this:

Dt[x Tan[y] - y^2 Log[x] == 4];
% /. {Dt[y] -> dy, Dt[x] -> dx};
Reduce[{%, dy/dx == z}, z]

On my system (Mathematica 9, Windows 7), this computation seems to hang.

Does anyone have a suggestion on how to solve this?


I'm also interested in why Reduce doesn't handle the problem well when used as above whereas Solve handles it just fine (see answers).

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up vote 7 down vote accepted

The command Dt accepts a second argument that specifies what symbol corresponds to the independent variable in the differentiation. In your case, that's x. The derivative you're looking for would then be written Dt[y, x].

So you just need to do this:

Solve[Dt[x Tan[y] - y^2 Log[x] == 4, x], Dt[y, x]]

==> {{Dt[y, x] -> (y^2 - x Tan[y])/(
   x (-2 y Log[x] + x Sec[y]^2))}}
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Thank you Jens. – dharmatech Aug 3 '13 at 23:38

You can also use D directly but define y as y[x] and solve for y'[x] as follows.

Solve[D[x Tan[y[x]] - y[x]^2 Log[x] == 4, x], y'[x]]
{{y'[x] -> (-x Tan[y[x]] + y[x]^2)/(
   x (x Sec[y[x]]^2 - 2 Log[x] y[x]))}}

And if you want the answer without the y[x] appearing you can use a replacement rule:

Solve[D[x Tan[y[x]] - y[x]^2 Log[x] == 4, x], y'[x]] /. y[x] -> y
{{y'[x] -> (y^2 - x Tan[y])/(
   x (-2 y Log[x] + x Sec[y]^2))}}
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Just as you started:

 expr1 = Dt[x Tan[y] - y^2 Log[x] == 4]

(* -((y^2 Dt[x])/x) - 2 y Dt[y] Log[x] + x Dt[y] Sec[y]^2 + 
  Dt[x] Tan[y] == 0  *)

Divide the both parts by Dt[x]:

    expr2 = Map[Divide[#, Dt[x]] &, expr1] // Apart

(* (-y^2 Dt[x] - 2 x y Dt[y] Log[x] + x^2 Dt[y] Sec[y]^2)/(x Dt[x]) + Tan[y] == 0 *)

Replace Dt[y]/Dt[x] by z:

 expr3 = expr2 /. Dt[y] -> z*Dt[x] // Simplify

  (*  x z Sec[y]^2 + Tan[y] == (y (y + 2 x z Log[x]))/x *)

Now find z:

    Solve[expr3, z]

(*  {{z -> (y^2 - x Tan[y])/(x (-2 y Log[x] + x Sec[y]^2))}}  *)


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