Question

1. xy - cos(xy) = 1

Explanation:

Step1: Differentiate both sides with respect to \(x\)

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\) and the chain rule \((f(g(x)))^\prime = f^\prime(g(x))\cdot g^\prime(x)\).
Differentiating \(xy\) gives \(y + x\frac{dy}{dx}\), differentiating \(-\cos(xy)\) gives \(\sin(xy)\cdot(y + x\frac{dy}{dx})\), and differentiating \(1\) gives \(0\). So we have:
\(y + x\frac{dy}{dx}+\sin(xy)\cdot(y + x\frac{dy}{dx})=0\)

Step2: Factor out \((y + x\frac{dy}{dx})\)

\((y + x\frac{dy}{dx})(1+\sin(xy)) = 0\)
Since \(1+\sin(xy)
eq0\) (because \(\sin(xy)\geq - 1\) and \(1+\sin(xy) = 0\) would imply \(\sin(xy)=-1\) which is a special case, but in general for non - singular points of the curve), we can divide both sides by \((1 + \sin(xy))\)
\(y+x\frac{dy}{dx}=0\)

Step3: Solve for \(\frac{dy}{dx}\)

\(x\frac{dy}{dx}=-y\)
\(\frac{dy}{dx}=-\frac{y}{x}\)

Answer:

\(\frac{dy}{dx}=-\frac{y}{x}\)