We have a function $x \sin y = y^2$. Let $P$ be $\displaystyle (\frac{\pi^2}{4}, \frac{\pi}{2})$. Evaluate $\displaystyle \frac{dx}{dy}$ and $\displaystyle \frac{d^2y}{dx^2}$ at $P$.
For $\displaystyle \frac{dx}{dy}$:
$\displaystyle \frac{\pi^2}{4} \sin \left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2$
$\displaystyle D_x \left[\frac{\pi^2}{4}\right] \sin \left(\frac{\pi}{2}\right) + \frac{\pi^2}{4} D_x \left[\sin \frac{\pi}{2}\right] = D_x \left[\left(\frac{\pi}{2}\right)^2\right]$
$\displaystyle \frac{\pi}{2} \sin \left(\frac{\pi}{2}\right) + \frac{\pi^2}{4} \cos \left(\frac{\pi}{2}\right) = \pi$
I'm wondering if I could directly substitute the points to the equation or solving first for $\frac{dx}{dy}$.
$\displaystyle x \sin y = y^2 \rightarrow x' \sin y + x D_x [\sin y] = D_x [y^2] \rightarrow \sin y + x \cos \frac{dx}{dy} = 2y$
I tried to use implicit differentiation. The left side by product rule and right by power rule.
Best Answer
We have $x \sin y = y^2$.
Taking the derivative with respect to $y$, we get $\displaystyle x \cos y + \frac{dx}{dy} \sin y = 2y$, and from this we get $\displaystyle \frac{dx}{dy} = \frac{2y - x \cos y}{\sin y}$.
Taking a derivative of the original equation with respect to $x$, we get $\sin y + x \cos y \frac{dy}{dx} = 2y \frac{dy}{dx}$, which means that $\displaystyle (2y - x \cos y) \frac{dy}{dx} = \sin y$, or $\displaystyle \frac{dy}{dx} = \frac{\sin y}{2y - x \cos y}$. Taking a derivative with respect to $x$ again, we have $\displaystyle \frac{d^2y}{dx^2} = \frac{\cos y \frac{dy}{dx} \cdot (2y - x \cos y) - \sin y (2 + x \sin y \frac{dy}{dx})}{(2y - x \cos y)^2}$. I'll let you do the substitution and simplification on your own.