How do you implicitly differentiate $22 = \frac{y}{1 - x e^{y}}$ $22=(y)/(1-xe^y)$ ?

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Answer 1 · 2018-04-07T19:00:23

$\frac{d y}{d x} = \frac{22 e^{y}}{y - 23}$ $(dy)/(dx)=(22e^y)/(y-23)$

We have;

$22 = \frac{y}{1 - x e^{y}}$ $22=y/(1-xe^y)$

$\Rightarrow 22 - 22 x e^{y} = y$ $=>22-22xe^y=y$

$22 - y = 22 x e^{y}$ $22-y=22xe^y$

$\frac{22 - y}{22 e^{y}} = x$ $(22-y)/(22e^y)=x$

$x = \frac{22 - y}{22 e^{y}}$ $x=(22-y)/(22e^y)$

Diff.w.r.t. $y$ $color(red)(y)$ and $"using"color(blue)" Quotient Rule"$

$(dx)/(dy)=(22e^yd/(dy)(22-y)-(22-y)d/(dy)(22e^y))/(22e^y)^2$

$=>(dx)/(dy)=(22e^y(-1)-(22-y)(22e^y))/(22e^y)^2$

$=>(dx)/(dy)=(22e^y(-1-22+y))/(22e^y)^2$

$=>color(red)((dx)/(dy))=(y-23)/(22e^y)to$ ,But , $[(dx)/(dy)=1/((dy)/(dx))]$

$=>color(red)((dy)/(dx))=(22e^y)/(y-23)$

How do you implicitly differentiate 22=(y)/(1-xe^y)22=y1−xey22=(y)/(1-xe^y)?