I am trying to find $\frac{\partial t}{ \partial y}$ as a function of $x$, $y$ and $\frac{\partial t}{\partial x}$ if both $x$ and $y$ depend on $t$ for the function

f[x_]=Tan[x*y] + E^y = -10*x 

but am uncertain as to how to do this. Any ideas or suggestions? Thank you in advance.


closed as off-topic by Jens, user9660, MarcoB, m_goldberg, rhermans Feb 20 '16 at 0:08

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  • 2
    $\begingroup$ You'll want to use Dt. You'll also want to compare the documentation for = and == to properly define equations. $\endgroup$ – IPoiler Feb 4 '16 at 22:30
  • $\begingroup$ D[Tan[xy] + Exp[y] + 10*x, t, NonConstants -> {x, y}] 10*D[x, t, NonConstants -> {x, y}] + Exp[y]*D[y, t, NonConstants -> {x, y}] + (yD[x, t, NonConstants -> {x, y}] + xD[y, t, NonConstants -> {x, y}])*(Sec[xy])^2 /. D[x, t, NonConstants -> {x, y}] -> x' Exp[y]*D[y, t, NonConstants -> {x, y}] + 10*x' + (Sec[x* y])^2*(xD[y, t, NonConstants -> {x, y}] + y x') /. D[y, t, NonConstants -> {x, y}] -> y' Solve[10*x' + Exp[y]*y' + (Sec[xy])^2*(yx' + x*y') == 0, y'] Is this correct? $\endgroup$ – Mary Feb 4 '16 at 22:50
  • $\begingroup$ How would I use Dt to solve this with respect to t? I have an exam on this tomorrow and would be SO grateful if you could explain! $\endgroup$ – Mary Feb 4 '16 at 22:56
  • 2
    $\begingroup$ The documentation explains explicitly how to use it. If you want the implicit derivative of an equation you simply apply it like a normal function, Dt[Tan[x*y]+E^y==-10x,t]. $\endgroup$ – IPoiler Feb 5 '16 at 2:10
  • $\begingroup$ If you have something to add to the question, please edit the question, code is very difficult to read in the comments, even worse if there is no effort to format it correctly. $\endgroup$ – rhermans Feb 20 '16 at 0:08