Find y' for e^xsin y - e^-ycos x = 0, and evaluate it

Find $y'$ for $e^{x}\sin y - e^{-y}\cos x = 0$, and evaluate it where $y = 0$.

Step 1: Find $x$ where $y = 0$

Setting $y = 0$ kills the first term, since $\sin 0 = 0$: \[0 - e^{0}\cos x = 0 \;\Longrightarrow\; \cos x = 0 \;\Longrightarrow\; x = \frac{\pi}{2}+k\pi.\]

There are infinitely many such points; we take the principal one $x = \dfrac{\pi}{2}$.

Step 2: Differentiate the first term

Product rule, with $\bigl(\sin y\bigr)' = y'\cos y$: $\frac{d}{dx}\bigl(e^{x}\sin y\bigr) = e^{x}\sin y + e^{x}\cos y\cdot y'$.

Step 3: Differentiate the second term

Product rule again, now with $\bigl(e^{-y}\bigr)' = -y'e^{-y}$ and $\bigl(\cos x\bigr)' = -\sin x$: $\frac{d}{dx}\bigl(e^{-y}\cos x\bigr) = -y'e^{-y}\cos x - e^{-y}\sin x$.

Step 4: Assemble and solve

Subtracting the second from the first and setting the result to zero, both minus signs flip: $e^{x}\sin y + e^{x}\cos y\cdot y' + e^{-y}\cos x\cdot y' + e^{-y}\sin x = 0$.

\[ y'\bigl(e^{x}\cos y + e^{-y}\cos x\bigr) = -\bigl(e^{x}\sin y + e^{-y}\sin x\bigr), \]

\[ y' = -\frac{e^{x}\sin y + e^{-y}\sin x}{e^{x}\cos y + e^{-y}\cos x}. \]

Step 5: Evaluate at $\left(\tfrac{\pi}{2},\,0\right)$

Here $\sin y = 0$, $\cos y = 1$, $\sin x = 1$, $\cos x = 0$ and $e^{-y} = 1$: $y' = -\frac{0+1}{e^{\pi/2}+0} = -e^{-\pi/2} \approx -0.20788$.

Step 6: The other branches

At a general point $x = \tfrac{\pi}{2}+k\pi$ with $y = 0$, the formula collapses to

\[y' = -\frac{\sin x}{e^{x}} = -\frac{(-1)^{k}}{e^{\pi/2+k\pi}}\]

\[k = 0:\ y' = -e^{-\pi/2} \approx -0.20788; \qquad k = 1:\ y' = +e^{-3\pi/2} \approx +0.00898 \;\;\checkmark\]

So the answer genuinely depends on which crossing is meant — the sign alternates and the magnitude decays exponentially.