Implicit differentiation and connected rates

What you’ll learn

• Differentiate a relationship containing both x and y.
• Find tangents to implicit curves.
• Connect rates of change through a shared variable.

Understand the idea

When $y$ depends on $x$, differentiating a term in $y$ requires the chain rule. For example, the derivative of $y^2$ is $2y(dy/dx)$. Differentiate both sides, collect all derivative terms and solve for $dy/dx$ before evaluating at a point. A product such as $xy$ requires the product rule.

Connected rates use the chain rule to link two quantities through an intermediate variable. If volume depends on radius and radius depends on time, then $dV/dt=(dV/dr)(dr/dt)$. Establish the geometric relationship first, differentiate it, then insert the instantaneous values and their units.

A verbal rate statement can define a differential equation: “increases at a rate proportional to its current amount” becomes $dy/dt=ky$ with $k>0$. Include signs, constants and initial conditions that express the model rather than treating every rate as positive.

Key formulae

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

$$\frac d{dx}(xy)=y+x\frac{dy}{dx}$$

$$\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}$$

$$\frac{dy}{dt}=ky\quad\text{for a rate proportional to }y$$

Worked example

Find the tangent gradient to $x^2+xy+y^2=7$ at $(1,2)$.

1. Differentiate: $2x+y+x(dy/dx)+2y(dy/dx)=0$.
2. Collect terms: $(x+2y)(dy/dx)=-(2x+y)$.
3. At $(1,2)$, the gradient is $-(2+2)/(1+4)=-4/5$.

Answer: Gradient $-4/5$; tangent $y-2=-\frac45(x-1)$.