Consider the curve $y^2+8x^2=1$. Find $\frac{dy}{dx}$ at the point $(0,-1)$

Expert intro: This curve is defined implicitly, so we differentiate both sides with respect to x and then substitute the point.

Detailed walkthrough

We are given the curve

$y^2+8x^2=1$

and asked to find $\frac{dy}{dx}$ at $(0,-1)$.

Step 1: Differentiate implicitly

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(y^2)+\frac{d}{dx}(8x^2)=\frac{d}{dx}(1)$

Using the chain rule on $y^2$:

$2y\frac{dy}{dx}+16x=0$

Step 2: Solve for $\frac{dy}{dx}$

$2y\frac{dy}{dx}=-16x$

$\frac{dy}{dx}=-\frac{16x}{2y}=-\frac{8x}{y}$

Step 3: Evaluate at $(0,-1)$

$\frac{dy}{dx}\bigg|_{(0,-1)}=-\frac{8\cdot 0}{-1}=0$

Answer

$\boxed{0}$

So the slope of the curve at $(0,-1)$ is $0$.

💡 Pitfall guide

A frequent error is differentiating $y^2$ as $2y$ only. Because $y$ depends on $x$, you must write $2y\frac{dy}{dx}$. Another common mistake is substituting the point before solving for $\frac{dy}{dx}$.

🔄 Real-world variant

If you were asked for the slope at a different point on the same curve, the derivative formula would stay $\frac{dy}{dx}=-\frac{8x}{y}$. You would only change the substitution step. If the curve were $y^2+ax^2=1$, the derivative would become $\frac{dy}{dx}=-\frac{ax}{y}$.

🔍 Related terms

implicit differentiation, derivative, slope