Question

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

Answer 1

Differentiate both sides implicitly:

$\dfrac{\mathrm d}{\mathrm dt}[e^y+x^3y^2+\ln x]=\dfrac{\mathrm d[1]}{\mathrm dt}$

$e^y\dfrac{\mathrm dy}{\mathrm dt}+3x^2y^2\dfrac{\mathrm dx}{\mathrm dt}+2x^3y\dfrac{\mathrm dy}{\mathrm dt}+\dfrac1x\dfrac{\mathrm dx}{\mathrm dt}=0$

Solve for $\frac{\mathrm dx}{\mathrm dt}$:

$\left(3x^2y^2+\dfrac1x\right)\dfrac{\mathrm dx}{\mathrm dt}=-(e^y+2x^3y)\dfrac{\mathrm dy}{\mathrm dt}$

$\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{e^y+2x^3y}{3x^2y^2+\frac1x}\dfrac{\mathrm dy}{\mathrm dt}$

$\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{xe^y+2x^4y}{3x^3y^2+1}\dfrac{\mathrm dy}{\mathrm dt}$

Plug in $x=1$, $y=0$, and $\frac{\mathrm dy}{\mathrm dt}=2$:

$\dfrac{\mathrm dx}{\mathrm dt}=\boxed{-2}$