How do you differentiate #1=e^(xy)/(e^x+xy)#?

Question

Paisley Johnson · Accepted Answer

#y'=(e^x+y-ye^x-xy^2)/(xe^x+x^2y-x)#

From the given

#1=e^(xy)/(e^x+xy)#

#e^x+xy=e^(xy)#

#d/dx(e^x)+d/dx(xy)=d/dx(e^(xy))#

#e^x*d/dx(x)+x*d/dx(y)+y*d/dx(x)=e^(xy)*[x*d/dx(y)+y*d/dx(x)]#

#e^x+xy'+y*1=e^(xy)*(xy'+y*1)#

#e^x+xy'+y=e^(xy)*(xy'+y)#

#e^x+y-y*e^(xy)=xy'*e^(xy)-xy'#

#e^x+y-y*e^(xy)=(x*e^(xy)-x)y'#

#y'=(e^x+y-y*e^(xy))/(x*e^(xy)-x)#

But, #e^(xy)=e^x+xy#

therefore

#y'=(e^x+y-y*e^(xy))/(x*e^(xy)-x)=(e^x+y-y*(e^x+xy))/(x*(e^x+xy)-x)#

#y'=(e^x+y-ye^x-xy^2)/(xe^x+x^2y-x)#

God bless....I hope the explanation is useful.

Evelyn Agard · Answer

undefined To differentiate $1 = \frac{e^{xy}}{e^x + xy}$ with respect to $x$, apply the quotient rule:

Let $u = e^{xy}$ and $v = e^x + xy$.

Then, apply the quotient rule:

$$
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}
$$

where $u'$ and $v'$ denote the derivatives of $u$ and $v$ with respect to $x$, respectively.

Now, find $u'$ and $v'$:

$$
u' = \frac{d}{dx}(e^{xy}) = ye^{xy}
$$

$$
v' = \frac{d}{dx}(e^x + xy) = e^x + y
$$

Substitute these into the quotient rule formula:

$$
\frac{d}{dx}\left(\frac{e^{xy}}{e^x + xy}\right) = \frac{(ye^{xy})(e^x + xy) - (e^{xy})(e^x + y)}{(e^x + xy)^2}
$$

Simplify this expression as needed.

Ezra Adams · Answer

undefined To differentiate $1 = \frac{e^{xy}}{e^x + xy}$, we use the quotient rule. The quotient rule states that for functions $u$ and $v$, the derivative of $u/v$ is given by:

$$
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
$$

In this case, let $u = e^{xy}$ and $v = e^x + xy$. We find the derivatives $\frac{du}{dx}$ and $\frac{dv}{dx}$:

$$
\frac{du}{dx} = \frac{d}{dx}(e^{xy}) = e^{xy} \cdot \frac{d(xy)}{dx} = ye^{xy}
$$

$$
\frac{dv}{dx} = \frac{d}{dx}(e^x + xy) = e^x + y
$$

Now, apply the quotient rule to find $\frac{d}{dx}\left(\frac{e^{xy}}{e^x + xy}\right)$:

$$
\frac{d}{dx}\left(\frac{e^{xy}}{e^x + xy}\right) = \frac{(e^x + xy)(ye^{xy}) - e^{xy}(e^x + y)}{(e^x + xy)^2}
$$

Simplify the numerator:

$$
= \frac{ye^{xy}(e^x + xy) - e^{xy}(e^x + y)}{(e^x + xy)^2}
$$

Expand and rearrange terms:

$$
= \frac{ye^{x + xy^2} - e^x e^{xy} - e^{xy}y}{(e^x + xy)^2}
$$

So, the derivative of $1 = \frac{e^{xy}}{e^x + xy}$ with respect to $x$ is:

$$
\frac{ye^{x + xy^2} - e^x e^{xy} - e^{xy}y}{(e^x + xy)^2}
$$