Implicit Differentiation — Question 9

Question 9 - Solution

We are given: $$x\sin(y) + y\cos(x) = 1$$

Differentiate both sides implicitly with respect to $x$: $$\frac{d}{dx}[x\sin(y)] + \frac{d}{dx}[y\cos(x)] = 0$$

Using the product rule: $$\sin(y) + x\cos(y)\frac{dy}{dx} + \cos(x)\frac{dy}{dx} - y\sin(x) = 0$$

Move the non-$\dfrac{dy}{dx}$ terms to the other side: $$x\cos(y)\frac{dy}{dx}+\cos(x)\frac{dy}{dx} = y\sin(x)-\sin(y)$$

Factor out $\dfrac{dy}{dx}$: $$\left(x\cos(y)+\cos(x)\right)\frac{dy}{dx} = y\sin(x)-\sin(y)$$

Solve for $\dfrac{dy}{dx}$: $$\boxed{ \frac{dy}{dx} = \frac{y\sin(x)-\sin(y)} {x\cos(y)+\cos(x)} }$$

At $(0,1)$: $$\frac{dy}{dx} = \frac{(1)\sin(0)-\sin(1)} {(0)\cos(1)+\cos(0)}$$

$$\frac{dy}{dx} = \frac{0-\sin(1)} {0+1} = \boxed{-\sin(1)}$$

Therefore, the slope of the tangent line at $(0,1)$ is: $$\boxed{-\sin(1)}$$