Find dy/dx for the relation sin(xy) + cos(x+y) = 1.

Question

ScanSolve AI Tutor · Accepted Answer

Here's the solution for question 8.

Differentiate both sides of the equation with respect to $x$ .
$\frac{d}{dx}(\sin(xy) + \cos(x+y)) = \frac{d}{dx}(1)$
Apply the chain rule and product rule to differentiate each term.
For $\sin(xy)$ :
Using the chain rule, $\frac{d}{dx}(\sin u) = \cos u \cdot \frac{du}{dx}$ . Here $u = xy$ .
Using the product rule, $\frac{d}{dx}(xy) = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x\frac{dy}{dx}$ .
So, $\frac{d}{dx}(\sin(xy)) = \cos(xy)\left(y + x\frac{dy}{dx}\right)$ .

For $\cos(x+y)$ :
Using the chain rule, $\frac{d}{dx}(\cos v) = -\sin v \cdot \frac{dv}{dx}$ . Here $v = x+y$ .
$\frac{d}{dx}(x+y) = 1 + \frac{dy}{dx}$ .
So, $\frac{d}{dx}(\cos(x+y)) = -\sin(x+y)\left(1 + \frac{dy}{dx}\right)$ .

For the right side:
$\frac{d}{dx}(1) = 0$ .
Substitute these derivatives back into the equation.
$\cos(xy)\left(y + x\frac{dy}{dx}\right) - \sin(x+y)\left(1 + \frac{dy}{dx}\right) = 0$
Expand the equation.
$y\cos(xy) + x\cos(xy)\frac{dy}{dx} - \sin(x+y) - \sin(x+y)\frac{dy}{dx} = 0$
Group terms containing $\frac{dy}{dx}$ on one side and other terms on the other side.
$x\cos(xy)\frac{dy}{dx} - \sin(x+y)\frac{dy}{dx} = \sin(x+y) - y\cos(xy)$
Factor out $\frac{dy}{dx}$ .
$\frac{dy}{dx}(x\cos(xy) - \sin(x+y)) = \sin(x+y) - y\cos(xy)$
Solve for $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \frac{\sin(x+y) - y\cos(xy)}{x\cos(xy) - \sin(x+y)}$
The final answer is $\frac{\sin(x+y) - y\cos(xy)}{x\cos(xy) - \sin(x+y)}$ .

Answer