Implicit and Parametric Differentiation (Further)

This lesson covers implicit differentiation and parametric differentiation at the Further Mathematics level. These techniques extend the chain rule to curves defined implicitly by an equation or parametrically in terms of a parameter.

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Implicit Differentiation

For a curve defined by $f(x, y) = 0$f(x,y)=0 (where $y$y is not given explicitly as a function of $x$x), differentiate both sides with respect to $x$x, treating $y$y as a function of $x$x and applying the chain rule.

The Chain Rule for Implicit Functions

When differentiating a term involving $y$y:

$\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}$dxd​[g(y)]=g′(y)⋅dxdy​

Worked Example 1: Find $\frac{dy}{dx}$dxdy​ for $x^2 + y^2 = 25$x2+y2=25 (a circle).

Differentiate both sides:

$2x + 2y\frac{dy}{dx} = 0$2x+2ydxdy​=0

$\frac{dy}{dx} = -\frac{x}{y}$dxdy​=−yx​

Worked Example 2: Find $\frac{dy}{dx}$dxdy​ for $x^3 + 3xy + y^3 = 1$x3+3xy+y3=1.

$3x^2 + 3\left(y + x\frac{dy}{dx}\right) + 3y^2\frac{dy}{dx} = 0$3x2+3(y+xdxdy​)+3y2dxdy​=0

$3x^2 + 3y + (3x + 3y^2)\frac{dy}{dx} = 0$3x2+3y+(3x+3y2)dxdy​=0

$\frac{dy}{dx} = -\frac{x^2 + y}{x + y^2}$dxdy​=−x+y2x2+y​

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Second Derivatives (Implicit)

For the second derivative, differentiate $\frac{dy}{dx}$dxdy​ again with respect to $x$x, using the quotient rule and substituting the expression for $\frac{dy}{dx}$dxdy​.