Revise Differentiation for Cambridge 9709 (3.4)

When we need to differentiate a function that is the product of two other functions (e.g., $x^2 \sin x$), we cannot simply differentiate each part separately. We must use the Product Rule. Similarly, for functions that are fractions (quotients) of other functions (e.g., $\frac{\ln x}{x}$), we use the Quotient Rule.

Product Rule: $y = uv \implies \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ \n Quotient Rule: $y = \frac{u}{v} \implies \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Sometimes, an equation connecting $x$ and $y$ cannot be easily rearranged to make $y$ the subject. For example, $x^3 + y^3 = 6xy$. Such equations define $y$ as an implicit function of $x$. To find the gradient $\frac{dy}{dx}$, we differentiate the entire equation term by term with respect to $x$, applying the chain rule whenever we differentiate a term involving $y$.

A curve can be described by expressing both $x$ and $y$ coordinates as functions of a third variable, or parameter, often denoted by $t$. For example, $x = 2t, y = t^2$. To find the gradient $\frac{dy}{dx}$ of such a curve, we use a version of the chain rule.

If $x=f(t)$ and $y=g(t)$, then $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

This formula allows us to find the gradient of the curve at any point by first finding the derivatives of $x$ and $y$ with respect to the parameter $t$, and then taking their ratio. This is often much simpler than trying to eliminate the parameter to get a Cartesian equation in $x$ and $y$.

In many real-world scenarios, several quantities vary with time and are related by some underlying formula. For example, as a spherical balloon is inflated, its volume, radius, and surface area all increase with time. Related rates problems involve finding the rate of change of one quantity, given the rate of change of another. The key is to use the chain rule with respect to time, $t$.