What is the slope of the tangent line to the curve sin(πx)+ln(x^2)y+sec(πx)=xy at x=1?

Question

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)

Answer 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 $D$.

Answer 2

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 $d)-(1+\pi).$

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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:

$-\pi-1$