Q.The equation of the normal to the parabola $y^2 = 8x$ at its origin is ________.

Implicit Differentiation

When $y$ isn't alone

You can differentiate $y=x^2+3x$ term by term because $y$ is written explicitly in terms of $x$. But an equation like $x^2+y^2=25$, or $x^3+y^3=6xy$, does not give $y$ by itself — solving for $y$ is messy or downright impossible.

Implicit differentiation finds $\dfrac{dy}{dx}$ without isolating $y$: treat $y$ as an unknown function of $x$, differentiate the whole equation as it stands, then solve for $\dfrac{dy}{dx}$.

The one key move: $y$ is really $y(x)$

Wherever $y$ appears, picture $y(x)$ hiding inside. Differentiating a $y$-term therefore needs the chain rule, which tacks on a factor of $\dfrac{dy}{dx}$:

That extra $\dfrac{dy}{dx}$ on every $y$-term is the whole trick.

The procedure

Worked example

For $x^2+y^2=25$:

The answer naturally contains both $x$ and $y$ — that is normal here. To get the slope at a point on the curve, substitute the coordinates after differentiating; there is no need to solve for $y$ first. …

Related Rates

The idea: quantities that change together

Many real situations involve two or more quantities that all vary with time, linked by a fixed relationship. Inflate a balloon and its radius and volume both grow; slide a ladder down a wall and the top's height and the foot's distance both change. A related-rates problem gives you the rate at which one quantity is changing and asks for the rate of another, at some instant.

The key insight: if the quantities are tied together by an equation, then their rates are tied together too. We uncover that link by differentiating the equation with respect to time $t$.

The core mechanism: differentiate with respect to time

Every variable is a function of $t$, so differentiating brings in the chain rule — each variable's derivative picks up a factor of its own rate. For example, if the volume of a sphere is $V=\tfrac{4}{3}\pi r^{3}$, then differentiating both sides with respect to $t$ gives

This single equation connects the rate the volume grows, $\dfrac{dV}{dt}$, to the rate the radius grows, $\dfrac{dr}{dt}$. Knowing one (and the current $r$) gives the other.

The standard procedure

Worked example

Air is pumped into a spherical balloon at $\dfrac{dV}{dt}=100\ \text{cm}^3/\text{s}$. How fast is the radius increasing when $r=5\ \text{cm}$?

From $\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}$, substitute $\dfrac{dV}{dt}=100$ and $r=5$:

$100 = 4\pi(5)^2\,\frac{dr}{dt} = 100\pi\,\frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{\pi}\ \text{cm/s}.$ …