1. Identify inner ($u = x^2$) and outer ($y = \sin(u)$) functions. 2. Differentiate: $\frac{dy}{du} = \cos(u)$, $\frac{du}{dx} = 2x$. 3. Apply chain rule: $\frac{dy}{dx} = \cos(x^2) \cdot 2x = 2x\cos(x^2)$.

1. Differentiate both sides: $2x + 2y\frac{dy}{dx} = 0$. 2. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{x}{y}$.

1. Find the first derivative, $f'(x)$. 2. Find the derivative of $f'(x)$ to get the second derivative, $f''(x)$. 3. Repeat to find higher derivatives.

Differentiate the outer function, evaluated at the inner function, multiplied by the derivative of the inner function.

Differentiate both sides of the equation with respect to $x$, applying the chain rule to terms involving $y$, and solve for $\frac{dy}{dx}$.

The concavity of the original function and helps find points of inflection.

The slope of the original function.

The derivative at a point is the slope of the tangent line at that point.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{d}{dx}(y^n) = n \cdot y^{n-1} \cdot \frac{dy}{dx}$

$\frac{f(b) - f(a)}{b - a}$