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Revision:Implicit Differentiation

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Implicit Differentiation

If y^3 = x, how would you differentiate this with respect to x? There are three ways:

Contents

Way One

Rewrite it as y = x^{\frac{1}{3}} and differentiate as normal.

Way Two

\displaystyle \frac{dx}{dy} = 3y^2


Since \displaystyle \frac{dy}{dx} =\frac{1}{\frac{dx}{dy}}

\displaystyle \frac{dy}{dx} = \frac{1}{3y^2}

Way Three

Differentiate term by term and use the chain rule:

\displaystyle y^3 = x

\displaystyle \frac{d (y^3)}{dx} = \frac{d (x)}{dx}


The right hand side of this equation is 1, since the derivative of x is 1. However, to work out the left hand side we must use the chain rule.


The left hand side becomes:

\displaystyle \frac{d (y^3)}{dy} \times \frac{dy}{dx}


(although it is not correct to do so, at this level you can think of dy/dx as a fraction in the chain rule. In the line above, imagine that you can cancel the 'dy' s, leaving d/dx and y³, which is what we had in the previous line).


Therefore,

\displaystyle 3y^2 \times \frac{dy}{dx} = 1

So

\displaystyle \frac{dy}{dx} = \frac{1}{3y^3}


In this example, method (2) is clearly the easiest. However, there are cases when the only possible method is (3).

Example

Differentiate \displaystyle x^2 + y^2 = 3x, with respect to x.


\displaystyle \frac{d (x^2)}{dx} + \frac{d (y^2)}{dx} = \frac{d (3x)}{dx}

\displaystyle 2x + \frac{d (y^2)}{dy} \times \frac{dy}{dx} = 3

\displaystyle 2x + 2y \frac{dy}{dx} = 3

\displaystyle \frac{dy}{dx} = \frac{3 - 2x}{2y}


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