Implicit Differentiation — Question 3

Question 3

Consider the equation: $$x^2\sin(y)+y^2\cos(x)=1$$

• Use implicit differentiation to find $\dfrac{dy}{dx}$.

• Find the slope of the tangent line to the curve at the point $(0,1)$.

Question 3 - Solution

We are given: $$x^2\sin(y)+y^2\cos(x)=1$$

Differentiate both sides implicitly with respect to $x$: $$\frac{d}{dx}\left[x^2\sin(y)\right] + \frac{d}{dx}\left[y^2\cos(x)\right] = 0$$

Using the product rule and chain rule: $$2x\sin(y)+x^2\cos(y)\frac{dy}{dx} + 2y\frac{dy}{dx}\cos(x)-y^2\sin(x) = 0$$

Move the non-$\dfrac{dy}{dx}$ terms to the other side: $$x^2\cos(y)\frac{dy}{dx} + 2y\cos(x)\frac{dy}{dx} = y^2\sin(x)-2x\sin(y)$$

Factor out $\dfrac{dy}{dx}$: $$\left(x^2\cos(y)+2y\cos(x)\right)\frac{dy}{dx} = y^2\sin(x)-2x\sin(y)$$

Solve for $\dfrac{dy}{dx}$: $$\boxed{ \frac{dy}{dx} = \frac{y^2\sin(x)-2x\sin(y)} {x^2\cos(y)+2y\cos(x)} }$$

At $(0,1)$: $$\frac{dy}{dx} = \frac{(1)^2\sin(0)-2(0)\sin(1)} {(0)^2\cos(1)+2(1)\cos(0)} = \frac{0}{2} = \boxed{0}$$

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