5.14 Implicit DifferentiationHL ONLY

1. Implicit Differentiation

So far, we have found derivatives of functions where $y$ is given explicitly in terms of $x$. For example, $y = x^2 + 3x$ or $y = e^x \sin x$.

However, sometimes the relationship between $y$ and $x$ is given implicitly, such as $x^2 + y^2 = 25$ or $e^y + xy = 10$. In these cases, it might be difficult or impossible to isolate $y$.

We use implicit differentiation. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$. This means whenever we differentiate a term involving $y$, we must apply the chain rule and multiply by $\dfrac{dy}{dx}$ (or $y'$).

EXAMPLE 1

Find $\dfrac{dy}{dx}$ if $x^2 + y^2 = 25$.

Solution:
Differentiating both sides with respect to $x$: $$ \begin{aligned} \dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(y^2) &= \dfrac{d}{dx}(25) \\ 2x + 2y \dfrac{dy}{dx} &= 0 \\ 2y \dfrac{dy}{dx} &= -2x \\ \dfrac{dy}{dx} &= \mathbf{-\dfrac{x}{y}} \end{aligned} $$ Notice that the derivative of $y^2$ is $2y \cdot \dfrac{dy}{dx}$ because of the chain rule.

EXAMPLE 2

Find the equation of the tangent line to the curve $x^3 + y^3 - 9xy = 0$ at the point $(2, 4)$.

Solution:
First, we differentiate implicitly to find $\dfrac{dy}{dx}$. Note that we must use the product rule for the term $9xy$: $$ \begin{aligned} 3x^2 + 3y^2 \dfrac{dy}{dx} - \left( 9(1)y + 9x \dfrac{dy}{dx} \right) &= 0 \\ 3x^2 + 3y^2 \dfrac{dy}{dx} - 9y - 9x \dfrac{dy}{dx} &= 0 \end{aligned} $$ Now, isolate the terms containing $\dfrac{dy}{dx}$: $$ \begin{aligned} 3y^2 \dfrac{dy}{dx} - 9x \dfrac{dy}{dx} &= 9y - 3x^2 \\ (3y^2 - 9x) \dfrac{dy}{dx} &= 9y - 3x^2 \\ \dfrac{dy}{dx} &= \dfrac{9y - 3x^2}{3y^2 - 9x} = \dfrac{3y - x^2}{y^2 - 3x} \end{aligned} $$ At the point $(2, 4)$, we evaluate the gradient $m$: $$ m = \dfrac{3(4) - (2)^2}{(4)^2 - 3(2)} = \dfrac{12 - 4}{16 - 6} = \dfrac{8}{10} = \dfrac{4}{5} $$ The equation of the tangent line is: $$ \begin{aligned} y - 4 &= \dfrac{4}{5}(x - 2) \\ y &= \mathbf{\dfrac{4}{5}x + \dfrac{12}{5}} \end{aligned} $$

2. More Kinematics

In Kinematics, acceleration $a = \dfrac{dv}{dt}$. By applying the chain rule, we can express acceleration in terms of velocity and displacement.

$a = \dfrac{dv}{dt} = \dfrac{dv}{ds} \cdot \dfrac{ds}{dt} = v \cdot \dfrac{dv}{ds}$

This formula is particularly useful when velocity or acceleration is given as a function of displacement ($s$) rather than time ($t$).

EXAMPLE 3

The velocity of a particle is given by $v = s^2 - 3s$, where $s$ is the displacement. Find the acceleration when $s = 4$.

Solution:
We use the formula $a = v \cdot \dfrac{dv}{ds}$.
First, we find $\dfrac{dv}{ds}$: $$ \dfrac{dv}{ds} = 2s - 3 $$ Thus, the acceleration is: $$ a = (s^2 - 3s)(2s - 3) $$ When $s = 4$: $$ \begin{aligned} a &= ((4)^2 - 3(4))(2(4) - 3) \\ &= (16 - 12)(8 - 3) \\ &= (4)(5) \\ &= \mathbf{20} \end{aligned} $$

If acceleration is given as a function of displacement $a = f(s)$, we can integrate to find velocity as a function of displacement:

$v \dfrac{dv}{ds} = a \implies \int v \, dv = \int a \, ds \implies \dfrac{1}{2}v^2 = \int a \, ds$

EXAMPLE 4

The acceleration of a particle is given by $a = 6s$, where $s$ is displacement. Initially, when $s = 0$, the velocity is $v = 4$. Find an expression for $v^2$ in terms of $s$.

Solution:
We use $\dfrac{1}{2}v^2 = \int a \, ds$: $$ \begin{aligned} \dfrac{1}{2}v^2 &= \int 6s \, ds \\ \dfrac{1}{2}v^2 &= 3s^2 + C \end{aligned} $$ Using the initial condition $s = 0$, $v = 4$: $$ \begin{aligned} \dfrac{1}{2}(4)^2 &= 3(0)^2 + C \\ 8 &= C \end{aligned} $$ Therefore: $$ \begin{aligned} \dfrac{1}{2}v^2 &= 3s^2 + 8 \\ v^2 &= \mathbf{6s^2 + 16} \end{aligned} $$