find y if 4 cos y / sin x = 1. ans) - cos x / 4 sin y ans) cos x / 4 sin y ans) - 4 sin y / cos x ans) 4 sin…

Explanation:

Step1: Differentiate both sides

Differentiate $\frac{\sin x}{4\cos y}=1$ with respect to $x$ using the quotient - rule and chain - rule. The quotient rule states that if $u = \sin x$ and $v = 4\cos y$, then $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$. The derivative of $\sin x$ with respect to $x$ is $\cos x$, and the derivative of $4\cos y$ with respect to $x$ is $- 4\sin y\cdot y'$ (by the chain - rule since $y$ is a function of $x$). So we have $\frac{\cos x\cdot4\cos y-\sin x\cdot(-4\sin y\cdot y')}{(4\cos y)^{2}} = 0$.

Step2: Simplify the equation

Multiply both sides by $(4\cos y)^{2}$ to get $\cos x\cdot4\cos y+\sin x\cdot4\sin y\cdot y'=0$. Then factor out 4: $4(\cos x\cos y+\sin x\sin y\cdot y') = 0$. Divide both sides by 4: $\cos x\cos y+\sin x\sin y\cdot y'=0$.

Step3: Solve for $y'$

Subtract $\cos x\cos y$ from both sides: $\sin x\sin y\cdot y'=-\cos x\cos y$. Then divide both sides by $\sin x\sin y$ (assuming $\sin x\neq0$ and $\sin y\neq0$) to get $y'=-\frac{\cos x\cos y}{\sin x\sin y}=-\cot x\cot y$. Another way is to rewrite the original equation $\sin x = 4\cos y$, and differentiate: $\cos x=-4\sin y\cdot y'$. Then $y'=-\frac{\cos x}{4\sin y}$.

Answer:

$-\frac{\cos x}{4\sin y}$