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

1 answer
Question:


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

Answers

Differentiate both sides implicitly:[tex]\dfrac{\mathrm d}{\mathrm dt}[e^y+x^3y^2+\ln x]=\dfrac{\mathrm d[1]}{\mathrm dt}[/tex][tex]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[/tex]Solve for [tex]\frac{\mathrm dx}{\mathrm dt}[/tex]:[tex]\left(3x^2y^2+\dfrac1x\right)\dfrac{\mathrm dx}{\mathrm dt}=-(e^y+2x^3y)\dfrac{\mathrm dy}{\mathrm dt}[/tex][tex]\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{e^y+2x^3y}{3x^2y^2+\frac1x}\dfrac{\mathrm dy}{\mathrm dt}[/tex][tex]\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{xe^y+2x^4y}{3x^3y^2+1}\dfrac{\mathrm dy}{\mathrm dt}[/tex]Plug in [tex]x=1[/tex], [tex]y=0[/tex], and [tex]\frac{\mathrm dy}{\mathrm dt}=2[/tex]:[tex]\dfrac{\mathrm dx}{\mathrm dt}=\boxed{-2}[/tex]

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