Find y' for xy - tan y = 0
Find $y'$ for $xy - \tan y = 0$.
Explanation of Method
A transcendental relation with no closed-form solution for y.
Step 1: Differentiate both sides with respect to $x$
Product rule on $xy$, chain rule on $\tan y$ with $\bigl(\tan y\bigr)' = \dfrac{y'}{\cos^{2}y}$: $y + xy' - \frac{y'}{\cos^{2}y} = 0$.
Step 2: Collect the terms containing $y'$
\[ y'\left(x - \frac{1}{\cos^{2}y}\right) = -y. \]
Step 3: Solve and clear the inner fraction
Multiplying numerator and denominator by $\cos^{2}y$: $y' = \frac{-y}{x-\dfrac{1}{\cos^{2}y}} = \frac{-y\cos^{2}y}{x\cos^{2}y-1}$.
Step 4: Verification
The relation $xy = \tan y$ is satisfied at $y = 0$ for every $x$, giving the trivial solution branch $y \equiv 0$: \[y = 0:\quad y' = -\frac{0\cdot 1}{x-1} = 0, \text{ consistent with the constant branch} \;\;\checkmark\]
\[x = 1:\quad \tan y = y \text{ has the root } y \approx 4.49341; \quad \cos^{2}y \approx 0.04718\]
\[y' \approx -\frac{4.49341\times 0.04718}{1\times 0.04718-1} \approx 0.22252 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = -\dfrac{y\cos^{2}y}{x\cos^{2}y-1}, \qquad x\cos^{2}y \neq 1\,}\]
Remark
Remark. The equation $\tan y = xy$ cannot be solved for $y$ in closed form — at $x = 1$ it is the classical transcendental equation $\tan y = y$, whose roots have no elementary expression. Implicit differentiation sidesteps this entirely: it gives the slope in terms of the coordinates of whatever point happens to lie on the curve.