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Higher Order and Implicit Differentiation

$$y' = \frac{dy}{dx}$$
$$y'' = \frac{d^2 y}{dx^2}$$
$$y^{(n)} = \frac{d^n y}{dx^n}$$

Implicit differentiation:

$$\frac{d}{dx}F(x,y) = F_x + F_y \frac{dy}{dx} = 0$$
$$\frac{dy}{dx} = -\frac{F_x}{F_y}$$

Steps in Solving Implicit Differentiation Problems

How to differentiate equations where $y$ is not isolated:

  1. Differentiate both sides of the equation with respect to $x$.
  2. Apply the chain rule whenever you differentiate a term involving $y$, which produces $\frac{dy}{dx}$.
  3. Collect all terms containing $\frac{dy}{dx}$ on one side of the equation.
  4. Factor out $\frac{dy}{dx}$.
  5. Solve for $\frac{dy}{dx}$.

Implicit differentiation is used when the equation cannot easily be solved for $y$, such as circles, ellipses, or equations where $x$ and $y$ are mixed together.

Slope of Tangent via Implicit Differentiation — CE Board

Find the slope of the tangent to the curve $x^2 + y^2 = 25$ at the point $(3,\,4)$.

Differentiate both sides implicitly with respect to $x$:

$2x + 2y\dfrac{dy}{dx} = 0$

Solve for $\dfrac{dy}{dx}$:

$\dfrac{dy}{dx} = -\dfrac{x}{y}$

Substitute $(3,\,4)$:

$\dfrac{dy}{dx}\bigg|_{(3,4)} = -\dfrac{3}{4}$

The slope of the tangent is $-\dfrac{3}{4}$.

Implicit Differentiation with Mixed Terms

Find $\dfrac{dy}{dx}$ given $x^2 + xy + y^2 = 7$.

Differentiate every term with respect to $x$ (apply product rule to $xy$):

$2x + \left(y + x\dfrac{dy}{dx}\right) + 2y\dfrac{dy}{dx} = 0$

Collect $\dfrac{dy}{dx}$ terms on one side:

$(x + 2y)\dfrac{dy}{dx} = -(2x + y)$
$\boxed{\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}}$

Implicit Differentiation — Trigonometric

Find $\dfrac{dy}{dx}$ if $\sin(x + y) = 2x$.

Differentiate both sides using the chain rule:

$\cos(x+y)\cdot\left(1 + \dfrac{dy}{dx}\right) = 2$

Solve for $\dfrac{dy}{dx}$:

$1 + \dfrac{dy}{dx} = \dfrac{2}{\cos(x+y)}$
$\boxed{\dfrac{dy}{dx} = \dfrac{2}{\cos(x+y)} - 1 = 2\sec(x+y) - 1}$

Equation of Tangent Line — CE Board Style

Find the equation of the tangent to the circle $x^2 + y^2 - 6x - 4y + 8 = 0$ at the point $(1,\,1)$.

Verify the point: $1 - 6 + 1 - 4 + 8 = 0$ ✓

Differentiate implicitly:

$2x + 2y\dfrac{dy}{dx} - 6 - 4\dfrac{dy}{dx} = 0$
$\dfrac{dy}{dx} = \dfrac{6 - 2x}{2y - 4} = \dfrac{3 - x}{y - 2}$

At $(1,\,1)$:

$m = \dfrac{3-1}{1-2} = \dfrac{2}{-1} = -2$

Using point-slope form:

$y - 1 = -2(x - 1)$
$\boxed{y = -2x + 3}$

Second-Order Implicit Derivative — CE Board

Given the curve $x^2 + y^2 = 25$, find the second derivative $\dfrac{d^2y}{dx^2}$.

Step 1: Find the first derivative.

$2x + 2y\,y' = 0 \implies y' = -\dfrac{x}{y}$

Step 2: Differentiate $y' = -\dfrac{x}{y}$ using the quotient rule.

$y'' = -\dfrac{(1)(y) - x\,y'}{y^2}$

Substitute $y' = -x/y$:

$y'' = -\dfrac{y - x\!\left(-\dfrac{x}{y}\right)}{y^2} = -\dfrac{y + \dfrac{x^2}{y}}{y^2} = -\dfrac{y^2 + x^2}{y^3}$

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

$\boxed{y'' = -\dfrac{25}{y^3}}$
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