MHB What is the equation of the tangent line at (π/2,1) for sin(xy)=y?

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The discussion focuses on finding the equation of the tangent line at the point (π/2, 1) for the implicit function sin(xy) = y. To solve this, implicit differentiation is applied, treating y as a function of x, leading to the equation cos(xy)(y + xy') = y'. After isolating y', the expression obtained is y' = -cos(xy)y / (x cos(xy) - 1). Substituting the point (π/2, 1) results in y' = 0, confirming that the tangent line at this point is y = 1.
karush
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Find the equation of the line tangent to
$$\sin\left({xy}\right)=y$$
At point
$$\left(\frac{\pi}{2 },1\right)$$
Answer $y=1$

I didn't know how to deal with xy.
No example given
 
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You need to differentiate implicitly, treating $y$ as $y(x)$. So, you'd start out with:
$$\d{}{x}\left[ \sin\left({xy}\right)=y \right] \qquad \implies \qquad \cos(xy) \cdot \left(y+xy'\right)=y'.$$
Can you continue from here?
 
Isolating $y'$ I got $$y '=\frac{-\cos\left({xy}\right)y}{x\cos\left({xy}\right)-1}$$
Plug in $$\left(\frac{\pi}{2 }, 1\right)\ \ y' =0$$
So $y=1$ is the equation
 
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