Question

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

Answer 1

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

Differentiate both sides, treating
`y` as a function of
`x`. 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
`y` 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))`