Differentiate: cosxy=x+3y Differentiate: cos xy = x + 3y
The Correct Answer and Explanation is:
We are asked to differentiate the implicit equation:cos(xy)=x+3y\cos(xy) = x + 3ycos(xy)=x+3y
To solve this, we will use implicit differentiation. The idea is to differentiate both sides of the equation with respect to xxx, treating yyy as an implicit function of xxx (i.e., y=y(x)y = y(x)y=y(x)).
Step 1: Differentiate the left side
The left-hand side involves the composite function cos(xy)\cos(xy)cos(xy). We’ll need to apply the chain rule here. The derivative of cos(u)\cos(u)cos(u) is −sin(u)-\sin(u)−sin(u), where u=xyu = xyu=xy, so we need to differentiate xyxyxy with respect to xxx.
Using the product rule to differentiate xyxyxy, we get:ddx(xy)=xdydx+y\frac{d}{dx}(xy) = x \frac{dy}{dx} + ydxd(xy)=xdxdy+y
Thus, the derivative of cos(xy)\cos(xy)cos(xy) with respect to xxx is:ddx[cos(xy)]=−sin(xy)⋅(xdydx+y)\frac{d}{dx} \left[ \cos(xy) \right] = -\sin(xy) \cdot \left( x \frac{dy}{dx} + y \right)dxd[cos(xy)]=−sin(xy)⋅(xdxdy+y)
Step 2: Differentiate the right side
The right-hand side of the equation is x+3yx + 3yx+3y. We differentiate each term:ddx(x)=1andddx(3y)=3dydx\frac{d}{dx}(x) = 1 \quad \text{and} \quad \frac{d}{dx}(3y) = 3 \frac{dy}{dx}dxd(x)=1anddxd(3y)=3dxdy
So, the derivative of the right-hand side is:1+3dydx1 + 3 \frac{dy}{dx}1+3dxdy
Step 3: Combine the derivatives
Now, we can equate the derivatives of both sides:−sin(xy)⋅(xdydx+y)=1+3dydx-\sin(xy) \cdot \left( x \frac{dy}{dx} + y \right) = 1 + 3 \frac{dy}{dx}−sin(xy)⋅(xdxdy+y)=1+3dxdy
Step 4: Solve for dydx\frac{dy}{dx}dxdy
To isolate dydx\frac{dy}{dx}dxdy, distribute −sin(xy)-\sin(xy)−sin(xy) on the left side:−sin(xy)⋅xdydx−sin(xy)⋅y=1+3dydx-\sin(xy) \cdot x \frac{dy}{dx} – \sin(xy) \cdot y = 1 + 3 \frac{dy}{dx}−sin(xy)⋅xdxdy−sin(xy)⋅y=1+3dxdy
Now, collect the terms involving dydx\frac{dy}{dx}dxdy on one side:−sin(xy)⋅xdydx−3dydx=1+sin(xy)⋅y-\sin(xy) \cdot x \frac{dy}{dx} – 3 \frac{dy}{dx} = 1 + \sin(xy) \cdot y−sin(xy)⋅xdxdy−3dxdy=1+sin(xy)⋅y
Factor out dydx\frac{dy}{dx}dxdy:(−sin(xy)⋅x−3)dydx=1+sin(xy)⋅y\left( -\sin(xy) \cdot x – 3 \right) \frac{dy}{dx} = 1 + \sin(xy) \cdot y(−sin(xy)⋅x−3)dxdy=1+sin(xy)⋅y
Finally, solve for dydx\frac{dy}{dx}dxdy:dydx=1+sin(xy)⋅y−sin(xy)⋅x−3\frac{dy}{dx} = \frac{1 + \sin(xy) \cdot y}{-\sin(xy) \cdot x – 3}dxdy=−sin(xy)⋅x−31+sin(xy)⋅y
Conclusion
The derivative of yyy with respect to xxx is:dydx=1+sin(xy)⋅y−sin(xy)⋅x−3\frac{dy}{dx} = \frac{1 + \sin(xy) \cdot y}{-\sin(xy) \cdot x – 3}dxdy=−sin(xy)⋅x−31+sin(xy)⋅y
This gives the rate of change of yyy with respect to xxx in terms of both xxx and yyy.
