$$\text { Find } \frac { d y } { d x }$$
-$\ln 5 x y = e ^ { x + y }$
A) $\frac { x y e ^ { x + y } - y } { x - x y e ^ { x + y } }$
B) $\frac { 2 x y e ^ { x + y } } { x + y }$
C) $\frac { y } { x }$
D) $\frac { e ^ { x + y } } { e ^ { 5 x } }$

Question

Quizplus · Accepted Answer

First, we need to use logarithmic differentiation to find the derivative of the function. Taking the natural logarithm of both sides:
$$\ln(\ln 5xy) = x + y$$
Differentiating implicitly with respect to $x$:
$$\frac{1}{\ln 5xy} \cdot \frac{d}{dx}(\ln 5xy) = 1 + \frac{dy}{dx}$$
Using the chain rule, we get:
$$\frac{1}{\ln 5xy} \cdot \frac{d}{dx}(\ln 5xy) = \frac{1}{5xy}\cdot (5y + 5xy') = 1 + \frac{dy}{dx}$$
Simplifying the left-hand side:
$$\frac{1}{\ln 5xy} \cdot \frac{5y + 5xy'}{5xy} = 1 + \frac{dy}{dx}$$
$$\frac{1}{\ln 5xy} \cdot \frac{y(1+x\frac{dy}{dx})}{xy} = 1 + \frac{dy}{dx}$$
Multiplying both sides by $\ln 5xy$:
$$\frac{5y + 5xy'}{5xy} = (\ln 5xy)(1 + \frac{dy}{dx})$$
$$\frac{y'}{y} + \frac{x'}{x} = (\ln 5xy)(1 + \frac{dy}{dx})$$
Substituting $x+y = \ln 5xy$, we can simplify the equation:
$$\frac{y'}{y} + \frac{x'}{x} = x+y'$$
Solving for $y'$:
$$y' = \frac{xye^{x+y} - y}{x - xye^{x+y}}$$
Therefore, the answer is $\boxed{	extbf{(A)}}$.

Find dydx\text { Find } \frac { d y } { d x } Find dxdy​

$\text { Find } \frac { d y } { d x }$