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How to find $\frac{dy}{dx}$ using implicit differentiation?

Differentiate both sides with respect to $x$ . 2. Apply chain rule to y terms. 3. Isolate $\frac{dy}{dx}$ .

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How to find $\frac{dy}{dx}$ using implicit differentiation?

Differentiate both sides with respect to $x$ . 2. Apply chain rule to y terms. 3. Isolate $\frac{dy}{dx}$ .

How to find the tangent line equation at a point?

Find $\frac{dy}{dx}$ at the point. 2. Use point-slope form: $y - y_1 = m(x - x_1)$ .

Explain the chain rule in implicit differentiation.

When differentiating a term involving $y$ , multiply by $\frac{dy}{dx}$ because $y$ is a function of $x$ .

Why is $\frac{dx}{dx} = 1$ ?

Because the rate of change of $x$ with respect to itself is always 1.

What does $\frac{dy}{dx}$ represent graphically?

The slope of the tangent line to the curve at a given point.

What is the general notation when differentiating both sides of an equation with respect to x?

$\frac{d}{dx}(\text{equation}) = \frac{d}{dx}(\text{other side})$

What is the point-slope form of a tangent line equation?

$y - y_1 = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is a point on the line.

State the product rule.

$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$