use implicit differentiation to find an equation of the tangent line to the curve at the given point. y…

Explanation:

Step1: Differentiate both sides

Differentiate $y\sin(16x)$ and $x\cos(2y)$ with respect to $x$ using product - rule and chain - rule. The product - rule states that $(uv)^\prime = u^\prime v+uv^\prime$. For $y\sin(16x)$: Let $u = y$ and $v=\sin(16x)$. Then $u^\prime=\frac{dy}{dx}$ and $v^\prime = 16\cos(16x)$. So, $(y\sin(16x))^\prime=\frac{dy}{dx}\sin(16x)+16y\cos(16x)$. For $x\cos(2y)$: Let $u = x$ and $v=\cos(2y)$. Then $u^\prime = 1$ and $v^\prime=- 2\sin(2y)\frac{dy}{dx}$. So, $(x\cos(2y))^\prime=\cos(2y)-2x\sin(2y)\frac{dy}{dx}$. We get $\frac{dy}{dx}\sin(16x)+16y\cos(16x)=\cos(2y)-2x\sin(2y)\frac{dy}{dx}$.

Step2: Solve for $\frac{dy}{dx}$

Rearrange the terms to isolate $\frac{dy}{dx}$: $\frac{dy}{dx}\sin(16x)+2x\sin(2y)\frac{dy}{dx}=\cos(2y)-16y\cos(16x)$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(\sin(16x)+2x\sin(2y))=\cos(2y)-16y\cos(16x)$. So, $\frac{dy}{dx}=\frac{\cos(2y)-16y\cos(16x)}{\sin(16x)+2x\sin(2y)}$.

Step3: Evaluate $\frac{dy}{dx}$ at the given point

Substitute $x = \frac{\pi}{2}$ and $y=\frac{\pi}{4}$ into $\frac{dy}{dx}$: $\sin(16x)=\sin(16\times\frac{\pi}{2})=\sin(8\pi)=0$. $\cos(16x)=\cos(8\pi)=1$. $\sin(2y)=\sin(2\times\frac{\pi}{4})=\sin(\frac{\pi}{2}) = 1$. $\cos(2y)=\cos(\frac{\pi}{2})=0$. $\frac{dy}{dx}\big|_{x = \frac{\pi}{2},y=\frac{\pi}{4}}=\frac{0 - 16\times\frac{\pi}{4}\times1}{0+2\times\frac{\pi}{2}\times1}=\frac{- 4\pi}{\pi}=-4$.

Step4: Find the equation of the tangent line

Use the point - slope form of a line $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(\frac{\pi}{2},\frac{\pi}{4})$ and $m=-4$. $y-\frac{\pi}{4}=-4(x - \frac{\pi}{2})$. $y-\frac{\pi}{4}=-4x + 2\pi$. $y=-4x+2\pi+\frac{\pi}{4}$. $y=-4x+\frac{8\pi+\pi}{4}$. $y=-4x+\frac{9\pi}{4}$.

Answer:

$y=-4x+\frac{9\pi}{4}$