Find dy/dx by implicit differentiation. cos(xy) = sin(x + y)

The Correct Answer and Explanation is:

To find dydx\frac{dy}{dx}dxdy by implicit differentiation of the equation cos⁡(xy)=sin⁡(x+y)\cos(xy) = \sin(x + y)cos(xy)=sin(x+y), follow these steps:

1. Differentiate both sides of the equation with respect to xxx.

We will apply the chain rule and product rule as necessary since both sides of the equation involve functions of xxx and yyy.

Left-hand side: cos⁡(xy)\cos(xy)cos(xy)

Differentiate cos⁡(xy)\cos(xy)cos(xy) with respect to xxx. Since the argument xyxyxy is a product of xxx and yyy, use the chain rule: ddx[cos⁡(xy)]=−sin⁡(xy)⋅ddx(xy)\frac{d}{dx} \left[ \cos(xy) \right] = -\sin(xy) \cdot \frac{d}{dx}(xy)dxd[cos(xy)]=−sin(xy)⋅dxd(xy)
To differentiate xyxyxy, apply the product rule: ddx(xy)=ddx(x)⋅y+x⋅ddx(y)=y+xdydx\frac{d}{dx}(xy) = \frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y) = y + x \frac{dy}{dx}dxd(xy)=dxd(x)⋅y+x⋅dxd(y)=y+xdxdy
Thus, the derivative of the left-hand side is: −sin⁡(xy)⋅(y+xdydx)-\sin(xy) \cdot (y + x \frac{dy}{dx})−sin(xy)⋅(y+xdxdy)

Right-hand side: sin⁡(x+y)\sin(x + y)sin(x+y)

Differentiate sin⁡(x+y)\sin(x + y)sin(x+y) with respect to xxx using the chain rule: ddx[sin⁡(x+y)]=cos⁡(x+y)⋅ddx(x+y)\frac{d}{dx} \left[ \sin(x + y) \right] = \cos(x + y) \cdot \frac{d}{dx}(x + y)dxd[sin(x+y)]=cos(x+y)⋅dxd(x+y)
The derivative of x+yx + yx+y is: ddx(x+y)=1+dydx\frac{d}{dx}(x + y) = 1 + \frac{dy}{dx}dxd(x+y)=1+dxdy
So, the derivative of the right-hand side is: cos⁡(x+y)⋅(1+dydx)\cos(x + y) \cdot (1 + \frac{dy}{dx})cos(x+y)⋅(1+dxdy)

2. Set up the equation for implicit differentiation:

Now, equate the derivatives of the left and right sides:−sin⁡(xy)⋅(y+xdydx)=cos⁡(x+y)⋅(1+dydx)-\sin(xy) \cdot (y + x \frac{dy}{dx}) = \cos(x + y) \cdot (1 + \frac{dy}{dx})−sin(xy)⋅(y+xdxdy)=cos(x+y)⋅(1+dxdy)

3. Solve for dydx\frac{dy}{dx}dxdy:

To isolate dydx\frac{dy}{dx}dxdy, expand both sides:−sin⁡(xy)⋅y−sin⁡(xy)⋅xdydx=cos⁡(x+y)⋅1+cos⁡(x+y)⋅dydx-\sin(xy) \cdot y – \sin(xy) \cdot x \frac{dy}{dx} = \cos(x + y) \cdot 1 + \cos(x + y) \cdot \frac{dy}{dx}−sin(xy)⋅y−sin(xy)⋅xdxdy=cos(x+y)⋅1+cos(x+y)⋅dxdy

Group the terms involving dydx\frac{dy}{dx}dxdy on one side:−sin⁡(xy)⋅xdydx−cos⁡(x+y)⋅dydx=cos⁡(x+y)+sin⁡(xy)⋅y-\sin(xy) \cdot x \frac{dy}{dx} – \cos(x + y) \cdot \frac{dy}{dx} = \cos(x + y) + \sin(xy) \cdot y−sin(xy)⋅xdxdy−cos(x+y)⋅dxdy=cos(x+y)+sin(xy)⋅y

Factor out dydx\frac{dy}{dx}dxdy:(−sin⁡(xy)⋅x−cos⁡(x+y))dydx=cos⁡(x+y)+sin⁡(xy)⋅y\left( -\sin(xy) \cdot x – \cos(x + y) \right) \frac{dy}{dx} = \cos(x + y) + \sin(xy) \cdot y(−sin(xy)⋅x−cos(x+y))dxdy=cos(x+y)+sin(xy)⋅y

Solve for dydx\frac{dy}{dx}dxdy:dydx=cos⁡(x+y)+sin⁡(xy)⋅y−sin⁡(xy)⋅x−cos⁡(x+y)\frac{dy}{dx} = \frac{\cos(x + y) + \sin(xy) \cdot y}{-\sin(xy) \cdot x – \cos(x + y)}dxdy=−sin(xy)⋅x−cos(x+y)cos(x+y)+sin(xy)⋅y

Final Answer:

dydx=cos⁡(x+y)+ysin⁡(xy)−xsin⁡(xy)−cos⁡(x+y)\frac{dy}{dx} = \frac{\cos(x + y) + y \sin(xy)}{-x \sin(xy) – \cos(x + y)}dxdy=−xsin(xy)−cos(x+y)cos(x+y)+ysin(xy)

This is the derivative of yyy with respect to xxx using implicit differentiation.