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

Question

Scarlett Allison · Accepted Answer

#(dy)/(dx)=(22e^y)/(y-23)#

We have;

#22=y/(1-xe^y)#

#=>22-22xe^y=y#

#22-y=22xe^y#

#(22-y)/(22e^y)=x#

#x=(22-y)/(22e^y)#

Diff.w.r.t. #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)#

Aurora Alvarado · Answer

undefined To implicitly differentiate $22 = \frac{y}{1-xe^y}$, we differentiate both sides of the equation with respect to $x$ using the chain rule and implicit differentiation techniques.

First, differentiate $22$ with respect to $x$ which results in $0$.

Then, differentiate $\frac{y}{1-xe^y}$ with respect to $x$.

The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$.

For the denominator, $1-xe^y$, differentiate $1-xe^y$ with respect to $x$, resulting in $-e^y\frac{dy}{dx}$.

Now apply the quotient rule, where $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:

$$
0 = \frac{-e^y\frac{dy}{dx} \cdot y - (1-xe^y) \frac{dy}{dx}}{(1-xe^y)^2}
$$

Rearrange and simplify to solve for $\frac{dy}{dx}$:

$$
(1-xe^y)\frac{dy}{dx} = -e^y\frac{dy}{dx} \cdot y
$$

$$
(1-xe^y + e^y y)\frac{dy}{dx} = 0
$$

$$
\frac{dy}{dx} = 0 \, \text{or} \, 1-xe^y + e^y y = 0
$$

Thus, the implicit differentiation of $22 = \frac{y}{1-xe^y}$ with respect to $x$ yields either $\frac{dy}{dx} = 0$ or $1-xe^y + e^y y = 0$.