{e}^(y) + {x}^(3) {y}^(2) + ln(x) = 1 and (dy)/(dt) = 2 when x=1 and y=0. Find (dx)/(dt) when x=1 and y=0​

Question

`{e}^(y) + {x}^(3) {y}^(2) + ln(x) = 1`

and

`(dy)/(dt) = 2`
when x=1 and y=0.
Find

`(dx)/(dt)`
when x=1 and y=0​

Answer 1

Differentiate both sides implicitly:

`(\mathrm d)/(\mathrm dt)[e^y+x^3y^2+\ln x]=(\mathrm d[1])/(\mathrm dt)`

`e^y(\mathrm dy)/(\mathrm dt)+3x^2y^2(\mathrm dx)/(\mathrm dt)+2x^3y(\mathrm dy)/(\mathrm dt)+\frac1x(\mathrm dx)/(\mathrm dt)=0`

Solve for
`(\mathrm dx)/(\mathrm dt)`:

`\left(3x^2y^2+\frac1x\right)(\mathrm dx)/(\mathrm dt)=-(e^y+2x^3y)(\mathrm dy)/(\mathrm dt)`

`(\mathrm dx)/(\mathrm dt)=-(e^y+2x^3y)/(3x^2y^2+\frac1x)(\mathrm dy)/(\mathrm dt)`

`(\mathrm dx)/(\mathrm dt)=-(xe^y+2x^4y)/(3x^3y^2+1)(\mathrm dy)/(\mathrm dt)`

Plug in
`x=1`,
`y=0`, and
`(\mathrm dy)/(\mathrm dt)=2`:

`(\mathrm dx)/(\mathrm dt)=\boxed{-2}`