Implicit Differentiation — Question 7

Question 7 - Solution

We are given: $$\sin(xy) + x^2 = y$$

(a) Differentiate both sides implicitly with respect to $x$:

Use the chain rule on $\sin(xy)$, noting that $xy$ is a product of functions: $$\frac{d}{dx}[\sin(xy)] + \frac{d}{dx}[x^2] = \frac{d}{dx}[y]$$

Differentiate: $$\cos(xy) \cdot \left( y + x \cdot \frac{dy}{dx} \right) + 2x = \frac{dy}{dx}$$

Now solve for $\frac{dy}{dx}$. Expand the left-hand side: $$\cos(xy)y + \cos(xy)x \cdot \frac{dy}{dx} + 2x = \frac{dy}{dx}$$

Bring all $\frac{dy}{dx}$ terms to one side: $$\cos(xy)x \cdot \frac{dy}{dx} - \frac{dy}{dx} = -\cos(xy)y - 2x$$

Factor $\frac{dy}{dx}$: $$\left( \cos(xy)x - 1 \right) \cdot \frac{dy}{dx} = -\cos(xy)y - 2x$$

Solve: $$\boxed{ \frac{dy}{dx} = \frac{-\cos(xy)y - 2x}{\cos(xy)x - 1} }$$

(b) Evaluate at the point $(0, 0)$:

Substitute $x = 0$, $y = 0$ into the derivative: $$\cos(0) = 1$$

$$\frac{dy}{dx} = \frac{ -1(0) - 2(0) }{ 1(0) - 1 } = \frac{0}{-1} = 0$$

$$\boxed{ \left. \frac{dy}{dx} \right|_{(0, 0)} = 0 }$$