Question

1. -/1 points use implicit differentiation to find an equation of the tangent line to the curve at the given point. y sin(12x) = x cos(2y), (π/2, π/4)

Explanation:

Step1: Differentiate both sides

Differentiate $y\sin(12x)=x\cos(2y)$ with respect to $x$ using product - rule and chain - rule.
The product - rule states that $(uv)^\prime = u^\prime v+uv^\prime$.
For the left - hand side:
The derivative of $y\sin(12x)$ with respect to $x$ is $y^\prime\sin(12x)+y\cdot12\cos(12x)$ (using $(uv)^\prime = u^\prime v + uv^\prime$ where $u = y$ and $v=\sin(12x)$).
For the right - hand side:
The derivative of $x\cos(2y)$ with respect to $x$ is $\cos(2y)+x(- 2\sin(2y)y^\prime)$ (using $(uv)^\prime=u^\prime v + uv^\prime$ where $u = x$ and $v=\cos(2y)$).
So we have $y^\prime\sin(12x)+12y\cos(12x)=\cos(2y)-2x\sin(2y)y^\prime$.

Step2: Solve for $y^\prime$

Rearrange the terms to isolate $y^\prime$:
$y^\prime\sin(12x)+2x\sin(2y)y^\prime=\cos(2y)-12y\cos(12x)$.
Factor out $y^\prime$: $y^\prime(\sin(12x)+2x\sin(2y))=\cos(2y)-12y\cos(12x)$.
Then $y^\prime=\frac{\cos(2y)-12y\cos(12x)}{\sin(12x)+2x\sin(2y)}$.

Step3: Evaluate $y^\prime$ at the given point

Substitute $x = \frac{\pi}{2}$ and $y=\frac{\pi}{4}$ into $y^\prime$:
$\sin(12x)=\sin(12\times\frac{\pi}{2})=\sin(6\pi)=0$,
$\cos(12x)=\cos(6\pi)=1$,
$\sin(2y)=\sin(2\times\frac{\pi}{4})=\sin(\frac{\pi}{2}) = 1$,
$\cos(2y)=\cos(\frac{\pi}{2})=0$.
$y^\prime=\frac{0 - 12\times\frac{\pi}{4}\times1}{0+2\times\frac{\pi}{2}\times1}=\frac{- 3\pi}{\pi}=-3$.

Step4: Find the equation of the tangent line

The equation of a line in point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(\frac{\pi}{2},\frac{\pi}{4})$ and $m=-3$.
$y-\frac{\pi}{4}=-3(x - \frac{\pi}{2})$.
Expand: $y-\frac{\pi}{4}=-3x+\frac{3\pi}{2}$.
Solve for $y$: $y=-3x+\frac{3\pi}{2}+\frac{\pi}{4}=-3x+\frac{6\pi + \pi}{4}=-3x+\frac{7\pi}{4}$.

Answer:

$y=-3x+\frac{7\pi}{4}$