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

Question

Use implicit differentiation to obtain $\dfrac {dy}{dx}$. $$x^2+y^2=16$$.

My Attempt: $$x^2+y^2=16$$ Differentiating both sides with respect to $x$, $$\dfrac {d}{dx} x^2 + \dfrac {d}{dx} y^2 = \dfrac {d}{dx} 16$$ $$2x+.....$$

Answer 1

if we assume, that $y=y(x)$ we get by the chain rule $$2x+2yy'=0$$

Answer 2

To implicitly differentiate, you must apply the chain rule:

$$\frac{\rm d}{\rm dx}(x^2+y^2)=\frac{\rm d}{\rm dx}(16)$$ $$2x+2y\frac{\rm dy}{\rm dx}=0$$ $$\frac{\rm dy}{\rm dx}=-\frac{x}{y}$$

Answer 3

Use the chain rule. Here you are assuming $y$ is a function of $x$ i.e you have $y = y(x)$. It now follows that, $y^2 = (y(x))^2 = x^2 \circ y(x)$ and so by the chain rule,

$$ \frac{d y^2}{dx} = \frac{d}{dx}(x^2) \cdot \frac{d}{dx}(y(x)) = 2x \cdot \frac{d y}{dx}$$