Answer:
E
Step-by-step explanation:
We are given:
[tex]\ln(xy)=x-y[/tex]
And we want to find the second derivative at the point (1, 1).
So, we will take the derivative of both sides with respect to x:
[tex]\displaystyle \frac{d}{dx}\big[\ln(xy)\big]=\frac{d}{dx}\big[x-y\big][/tex]
Implicitly differentiate. The left will require the chain rule. Hence:
[tex]\displaystyle \frac{1}{xy}\cdot \frac{d}{dx}\big[xy\big]=1-\frac{dy}{dx}[/tex]
Differentiate:
[tex]\displaystyle \frac{1}{xy}\cdot \big( y+x \frac{dy}{dx} \big) =1-\frac{dy}{dx}[/tex]
Distribute:
[tex]\displaystyle \frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}[/tex]
Isolate the derivative term:
[tex]\displaystyle \frac{dy}{dx}\big(\frac{1}{y}+1\big)=1-\frac{1}{x}[/tex]
Divide:
[tex]\displaystyle \frac{dy}{dx}=\frac{1-\frac{1}{x}}{\frac{1}{y}+1}}[/tex]
Simplify. We can multiply both layers by xy. Hence:
[tex]\displaystyle\begin{aligned} \frac{dy}{dx}&=\frac{1-\frac{1}{x}}{\frac{1}{y}+1}}\cdot\frac{xy}{xy} \\ &=\frac{xy-y}{x+xy} \end{aligned}[/tex]
We can factor. Hence, our derivative is:
[tex]\displaystyle \frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}[/tex]
Differentiate once more to find the second derivative:
[tex]\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\Big[\frac{y(x-1)}{x(y+1)}\Big][/tex]
Quotient rule:
[tex]\displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dx}[y(x-1)](x(y+1))-y(x-1)\frac{d}{dx}[x(y+1)]}{(x(y+1))^2}[/tex]
Differentiate using the product rule:
[tex]\displaystyle \frac{d^2y}{dx^2}=\frac{ \big[\frac{dy}{dx}(x-1)+y\big](x(y+1))-y(x-1)\big[(y+1)+x\frac{dy}{dx}\big] }{(x(y+1))^2 }[/tex]
You may choose to simplify, but this is not necessary, as we are only interested in the second derivative at (1, 1).
First, since first derivatives exist in our second derivative, we will find them first. Recall that the first derivative is given by:
[tex]\displaystyle \frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}[/tex]
Therefore, at (1, 1), the first derivative is:
[tex]\displaystyle \frac{dy}{dx}_{(1, 1)}=\frac{1(1-1)}{1(1+1)}=\frac{1(0)}{1(2)}=0[/tex]
So, we will substitute 0 for every dy/dx for our second derivative. Thus:
[tex]\displaystyle \frac{d^2y}{dx^2}=\frac{ \big[(0)(x-1)+y\big](x(y+1))-y(x-1)\big[(y+1)+x(0)\big] }{(x(y+1))^2 }[/tex]
Simplify:
[tex]\displaystyle \frac{d^2y}{dx^2}=\frac{ y(x(y+1))-y(x-1)\big[(y+1)\big] }{(x(y+1))^2 }[/tex]
So, at the point (1, 1), our second derivative is:
[tex]\displaystyle \frac{d^2y}{dx^2}_{(1, 1)}=\frac{1(1(1+1))-1(1-1)(1+1)}{(1(1+1))^2}=\frac{2-0}{4}=\frac{1}{2}[/tex]
Hence, our final answer is E.
