Questions: Find dy/dx by implicit differentiation. dy/dx= e^(x / y)=4 x-y Use implicit differentiation to find an equation of the tangent line to the curve at the given point. y sin (8 x)=x cos (2 y), (π / 2, π / 4)

Find dy/dx by implicit differentiation.
dy/dx= e^(x / y)=4 x-y

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
y sin (8 x)=x cos (2 y), (π / 2, π / 4)

Transcript text: Find $\frac{d y}{d x}$ by implicit differentiation. \[ \frac{d y}{d x}=\square e^{x / y}=4 x-y \] Use implicit differentiation to find an equation of the tangent line to the curve at the given point. \[ y \sin (8 x)=x \cos (2 y), \quad(\pi / 2, \pi / 4) \]

Solution

Solution Steps

To find the equation of the tangent line to the curve $ y \sin(8x) = x \cos(2y) $ at the point $(\pi/2, \pi/4)$, we need to follow these steps:

Use implicit differentiation to differentiate both sides of the equation with respect to $ x $.
Solve for $ \frac{dy}{dx} $ to find the slope of the tangent line at the given point.
Use the point-slope form of the equation of a line to write the equation of the tangent line.

Step 1: Differentiate both sides of the equation implicitly

Given the equation: \[ y \sin (8 x) = x \cos (2 y) \] we need to differentiate both sides with respect to $x$. Remember to use the product rule and chain rule where necessary.

Step 2: Apply the product rule and chain rule

Differentiate the left-hand side: \[ \frac{d}{dx} \left( y \sin (8 x) \right) = y \frac{d}{dx} \left( \sin (8 x) \right) + \sin (8 x) \frac{d y}{d x} \] Using the chain rule: \[ \frac{d}{dx} \left( \sin (8 x) \right) = 8 \cos (8 x) \] So, the left-hand side becomes: \[ y \cdot 8 \cos (8 x) + \sin (8 x) \frac{d y}{d x} \]

Differentiate the right-hand side: \[ \frac{d}{dx} \left( x \cos (2 y) \right) = \cos (2 y) + x \frac{d}{dx} \left( \cos (2 y) \right) \] Using the chain rule: \[ \frac{d}{dx} \left( \cos (2 y) \right) = -2 \sin (2 y) \frac{d y}{d x} \] So, the right-hand side becomes: \[ \cos (2 y) - 2 x \sin (2 y) \frac{d y}{d x} \]

Step 3: Combine and solve for $\frac{d y}{d x}$

Equate the differentiated expressions: \[ y \cdot 8 \cos (8 x) + \sin (8 x) \frac{d y}{d x} = \cos (2 y) - 2 x \sin (2 y) \frac{d y}{d x} \]

Rearrange to isolate $\frac{d y}{d x}$: \[ \sin (8 x) \frac{d y}{d x} + 2 x \sin (2 y) \frac{d y}{d x} = \cos (2 y) - y \cdot 8 \cos (8 x) \]

Factor out $\frac{d y}{d x}$: \[ \left( \sin (8 x) + 2 x \sin (2 y) \right) \frac{d y}{d x} = \cos (2 y) - y \cdot 8 \cos (8 x) \]

Solve for $\frac{d y}{d x}$: \[ \frac{d y}{d x} = \frac{\cos (2 y) - y \cdot 8 \cos (8 x)}{\sin (8 x) + 2 x \sin (2 y)} \]

Step 4: Evaluate $\frac{d y}{d x}$ at the given point $(\pi / 2, \pi / 4)$

Substitute $x = \pi / 2$ and $y = \pi / 4$ into the expression: \[ \frac{d y}{d x} \bigg|_{(\pi / 2, \pi / 4)} = \frac{\cos (2 \cdot \pi / 4) - (\pi / 4) \cdot 8 \cos (8 \cdot \pi / 2)}{\sin (8 \cdot \pi / 2) + 2 (\pi / 2) \sin (2 \cdot \pi / 4)} \]

Simplify the trigonometric functions: \[ \cos (\pi / 2) = 0, \quad \cos (4 \pi) = 1, \quad \sin (4 \pi) = 0, \quad \sin (\pi / 2) = 1 \]

Substitute these values: \[ \frac{d y}{d x} \bigg|_{(\pi / 2, \pi / 4)} = \frac{0 - (\pi / 4) \cdot 8 \cdot 1}{0 + 2 (\pi / 2) \cdot 1} = \frac{-2 \pi}{\pi} = -2 \]

Step 5: Find the equation of the tangent line

The slope of the tangent line at $(\pi / 2, \pi / 4)$ is $-2$. Use the point-slope form of the equation of a line: \[ y - y_1 = m (x - x_1) \] where $m = -2$, $x_1 = \pi / 2$, and $y_1 = \pi / 4$: \[ y - \frac{\pi}{4} = -2 \left( x - \frac{\pi}{2} \right) \]

Simplify: \[ y - \frac{\pi}{4} = -2x + \pi \] \[ y = -2x + \pi + \frac{\pi}{4} \] \[ y = -2x + \frac{5\pi}{4} \]