Find thd (dy)/(dx) : x³y²+sin(x㏑y)+e^(xy)=0

Question

asked Nov 9, 2021 65.8k views

1 Answer

Dexter Huinda · Answer 1 · 2021-11-15T09:24:45+0000

$x^3y^2+\sin(x\ln y)+e^(xy)=0$

Differentiate both sides, treating
as a function of
. Let's take it one term at a time.

Power, product and chain rules:

$(\mathrm d(x^3y^2))/(\mathrm dx)=(\mathrm d(x^3))/(\mathrm dx)y^2+x^3(\mathrm d(y^2))/(\mathrm dx)$

$=3x^2y^2+x^3(2y)(\mathrm dy)/(\mathrm dx)$

$=3x^2y^2+6x^3y(\mathrm dy)/(\mathrm dx)$

Product and chain rules:

$(\mathrm d(\sin(x\ln y))/(\mathrm dx)=\cos(x\ln y)(\mathrm d(x\ln y))/(\mathrm dx)$

$=\cos(x\ln y)\left((\mathrm d(x))/(\mathrm dx)\ln y+x(\mathrm d(\ln y))/(\mathrm dx)\right)$

$=\cos(x\ln y)\left(\ln y+\frac1y(\mathrm dy)/(\mathrm dx)\right)$

$=\cos(x\ln y)\ln y+\frac{\cos(x\ln y)}y(\mathrm dy)/(\mathrm dx)$

Product and chain rules:

$(\mathrm d(e^(xy)))/(\mathrm dx)=e^(xy)(\mathrm d(xy))/(\mathrm dx)$

$=e^(xy)\left((\mathrm d(x))/(\mathrm dx)y+x(\mathrm d(y))/(\mathrm dx)\right)$

$=e^(xy)\left(y+x(\mathrm dy)/(\mathrm dx)\right)$

$=ye^(xy)+xe^(xy)(\mathrm dy)/(\mathrm dx)$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y(\mathrm dy)/(\mathrm dx)+\cos(x\ln y)\ln y+\frac{\cos(x\ln y)}y(\mathrm dy)/(\mathrm dx)+ye^(xy)+xe^(xy)(\mathrm dy)/(\mathrm dx)=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\frac{\cos(x\ln y)}y+xe^(xy)\right)(\mathrm dy)/(\mathrm dx)=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^(xy)\right)$

$(\mathrm dy)/(\mathrm dx)=-\frac{3x^2y^2+\cos(x\ln y)\ln y-ye^(xy)}{6x^3y+\frac{\cos(x\ln y)}y+xe^(xy)}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by
to get rid of that fraction in the denominator.

$(\mathrm dy)/(\mathrm dx)=-(3x^2y^3+y\cos(x\ln y)\ln y-y^2e^(xy))/(6x^3y^2+\cos(x\ln y)+xye^(xy))$