17. What is the slope of the tangent to the curve \( \sin(\pi x) + 9\cos(\pi y) = x^2y \) at \( (3, -1) \)?






Answer is: option3

\( \frac{6 - \pi}{9} \)

Solution:

You use implicit differentiation to obtain:

\[ \pi \cos(\pi x) - 9\pi y' \sin(\pi y) = 2xy + x^2 y' \] \[ \pi \cos(\pi x) - 2xy = y'(9\pi \sin(\pi y) - x^2) \] \[ y' = \frac{\pi \cos(\pi x) - 2xy}{9\pi \sin(\pi y) - x^2} \]

When you evaluate this last expression at \( (3, -1) \) you obtain:

\[ \frac{6 - \pi}{9} Ans \]

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