sin(πx)+ln(x^2)y+sec(πx)=xy
How am I supposed to sub in x=1 if I don't know what y is? I got:
dy/dx = y-(2y/x)-πcos(πx)-πsec(πx)tan(πx)/(2ln(x)-x)
But then how do I solve this question if I got dy/dx or is that not what I was supposed to do?
Possible answers:
A: -2π
B: -π
C: -1
D: -(π+1)
Start by rearranging your equation as follows:
$xy-ln(x^2)y=sin(πx)+sec(πx)$
$y=\frac{sin(πx)+sec(πx)}{x-ln(x^2)}$.
Then it is straightforward to find the derivative and then substitute $x=1$.
The answer should be $d)-(1+\pi).$
More details:
$\frac{dy}{dx}=\frac{(x-ln(x^2))\frac{d}{dx} (sin(\pi x)+sec(\pi x))}{(x-ln(x^2))^2}$-$ \frac{ (sin(\pi x)+sec(\pi x))\frac{d}{dx}(x-ln(x^2))}{(x-ln(x^2))^2}$=$\frac{\pi \cos (\pi x)+\pi \tan (\pi x) \sec (\pi x)}{x-ln(x^2)}$-$ \frac{ (sin(\pi x)+sec(\pi x))(1-\frac{2}{x})}{(x-ln(x^2))^2}$
Then setting $x=1$, you get:
$-\pi-1$
$\sin (\pi x) + y \ln x^2 + \sec (\pi x) = xy$
By implicit differentiation, we get
$$\pi \cos (\pi x) + \frac{dy}{dx} \ln x^2 + \frac{2y}{x} + \pi \sec (\pi x) \tan (\pi x) = x \frac{dy}{dx} + y$$
Rearranging the equation by putting the $dy/dx$ term on the left side, we get
$$(2 \ln x - x) \frac{dy}{dx} = (1 - \frac{2}{x}) y - \pi (\cos (\pi x) + \sec (\pi x) \tan (\pi x))$$
We also note that, at $x = 1$, the original equation becomes $\sin \pi + y \ln 1 + \sec \pi = y$, or $y = -1$.
Plugging $x = 1$ and $y = -1$ into the equation, we get
$$(2 \ln 1 - 1) \frac{dy}{dx} = (1 - \frac{2}{1}) (-1) - \pi (\cos (\pi) + \sec (\pi) \tan (\pi))$$
or
$$-\frac{dy}{dx} = 1 + \pi$$
which means that the answer is $D$.
