Question

50 points each question. Please help. How do I solve?

Answer 1

Answer:

$\dfrac{dy}{dx}=-y\left(4+\dfrac{1}{x}\right)$

or written in rational form:

$\dfrac{dy}{dx}=\dfrac{-4xy-y}{x}$

Step-by-step explanation:

$\ln(xy)+4x=45$

$\textsf{Apply the product log law}\quad\ln(ab)=\ln(a)+\ln(b):$

$\implies \ln(x)+\ln(y)+4x=45$

$\textsf{Add}\:\dfrac{d}{dx}\:\textsf{in front of each term}:$

$\implies \dfrac{d}{dx}\ln(x)+\dfrac{d}{dx}\ln(y)+\dfrac{d}{dx}4x=\dfrac{d}{dx}45$

$\textsf{Differentiate the}\: x \: \textsf{terms and constant terms first}:$

$\implies \dfrac{1}{x}+\dfrac{d}{dx}\ln(y)+4=0$

$\textsf{When differentiating with respect to}\:y,\: \textsf{differentiate and add}\: \dfrac{dy}{dx}:$

$\implies \dfrac{1}{x}+\dfrac{1}{y}\dfrac{dy}{dx}+4=0$

$\textsf{Rearrange to make}\:\dfrac{dy}{dx}\:\textsf{the subject}:$

$\implies \dfrac{1}{y}\dfrac{dy}{dx}=-4-\dfrac{1}{x}$

$\implies \dfrac{dy}{dx}=y\left(-4-\dfrac{1}{x}\right)$

$\implies \dfrac{dy}{dx}=-y\left(4+\dfrac{1}{x}\right)$

If you want it in rational form, then:

$\implies \dfrac{dy}{dx}=-4y-\dfrac{y}{x}$

$\implies \dfrac{dy}{dx}=\dfrac{-4xy-y}{x}$

Answer 2

$~~~~~\ln(xy) +4x = 45\\\\\\\implies \dfrac{d}{dx} \left[\ln(xy) +4x \right] = \dfrac{d}{dx} ( 45)\\\\\\\implies \dfrac{1}{xy} \cdot \dfrac{d}{dx} (xy) + 4 = 0\\\\\\\implies \dfrac 1{xy} \left( x\dfrac{dy}{dx} + y \right) +4 =0\\\\\\\implies x\dfrac{dy}{dx} + y + 4xy=0\\\\\\\implies x \dfrac{dy}{dx} =-y-4xy\\\\\\\implies \dfrac{dy}{dx} = \dfrac{-y-4xy}x$