5FImplicitDiff

Implicit Differentiation

Our normal rules for differentiation require us to start with an equation in the form // y = f(x) //.

Some equations cannot be stated in this form:

math \text {Example: } \; xy + y^3 = x^2 math

This cannot be rearranged to make y the subject.

So we use __Implicit Differentiation__ (which relies on the Chain Rule)


 * Example 1 **

math \text{Find the derivative of } \; xy + y^3 = x^2 math


 * Solution:**

math . \qquad \dfrac {d}{dx} \left( xy+y^3 \right) = \dfrac {d}{dx} \left( x^2 \right) math
 * Take the derivative of both sides with respect to x

math . \qquad \dfrac {d}{dx} \left( xy \right) + \dfrac {d}{dx} \left( y^3 \right) = \dfrac {d}{dx} \left( x^2 \right) math

math . \qquad \Big( x \dfrac {dy}{dx} + y \dfrac {dx}{dx} \Big) + \dfrac{d \left( y^3 \right)}{dy} \dfrac {dy}{dx} = \dfrac {d}{dx} \left( x^2 \right) math . math . \qquad \Big( x \dfrac {dy}{dx} + y \dfrac {dx}{dx} \Big) + 3y^2 \dfrac {dy}{dx} = 2x math
 * Use the __ product rule __ on the first term and the __ chain rule __ on the second term

math . \qquad x \dfrac {dy}{dx} + y + 3y^2 \dfrac {dy}{dx} = 2x math
 * dx/dx=1 so simplify and rearrange to make dy/dx the subject

math . \qquad x \dfrac {dy}{dx} + 3y^2 \dfrac {dy}{dx} = 2x - y math

math . \qquad \dfrac {dy}{dx} \left( x + 3y^2 \right) = 2x - y math

math . \qquad \dfrac {dy}{dx} = \dfrac {2x - y}{x + 3y^2} math

Example 2

a) Use implicit differentiation to find the derivative of: math x^2 + y^2 = 9 math

b) and hence, find the gradient of the tangent to the curve at x = 1.


 * Solution:**

math . \qquad \dfrac {d}{dx} \left( x^2 \right) + \dfrac {d }{dx} \left( y^2 \right) = \dfrac {d}{dx} \left( 9 \right) math
 * Take the derivative of both sides with respect to x

math . \qquad 2x + \dfrac{d \left( y^2 \right) }{dy} \dfrac {dy}{dx} = 0 math . math . \qquad 2x + 2y \dfrac {dy}{dx} = 0 math
 * Differentiate each term, using chain rule where appropriate

math . \qquad 2y \dfrac {dy}{dx} = -2x math
 * Rearrange to make dy/dx the subject

math . \qquad \dfrac {dy}{dx} = \dfrac {-2x}{2y} = \dfrac {-x}{y} math

b) Hence find the gradient of the tangent to the curve at x = 1 {at x = 1} math . \qquad 1 + y^2 = 9 math

math . \qquad y^2 = 8 math

math . \qquad y = \pm \sqrt 8 = \pm 2 \sqrt 2 math

math . \qquad \dfrac {dy}{dx} = \dfrac {-x}{y} math

math . \qquad \quad = \dfrac {-1}{\pm 2 \sqrt 2} math

math . \qquad \quad = \dfrac {\pm \sqrt 2}{4} math

math . \qquad \dfrac {dy}{dx} = \dfrac {-x}{y} math
 * NOTE 1 **
 * In this particular example,

math . \qquad y = \pm \sqrt {9-x^2} math
 * can be simplified further because the original equation can be rearranged to make y the subject

Therefore math . \qquad \dfrac {dy}{dx} = \dfrac {-x}{\pm \sqrt{9-x^2}} = \dfrac {\pm x}{\sqrt{9-x^2}} math


 * NOTE 2 **

The derivative clearly has an implied domain because it involves dividing by a square root, so: math . \qquad 9-x^2 > 0 math

math . \qquad -x^2 > -9 math

math . \qquad x^2 < 9 \qquad \textit {dividing by a -ve reverses direction of inequality} math

math . \qquad |x| < 3 math

This means the derivative of x 2 + y 2 = 9 is not defined at the endpoints x = ±3 (as we would expect) but a tangent can be drawn to the circle at those points. (a vertical line, so gradient is infinity)

Implicit Differentiation on the Calculator

Your calculator can do implicit differentiation.
 * impDiff** is in the ACTION menu, CALCULATION submenu.

Example 1 Enter math . \qquad \text {impDiff} (xy + y^3 = x^2, x, y) math

Example 2 Enter math . \qquad \text {impDiff} (x^2 + y^2 = 9, x, y) math

The calculator shows the derivative as y'. You should write your answer using standard notation: math \text{Either: } f'(x) \text{ or } \dfrac{dy}{dx} \text{ or } y'(x) math (and don't write the dots)

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