Revision:Implicit Differentiation

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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:

1 Way One
2 Way Two
3 Way Three
4 Example
5 Comments

Way One

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

Way Two

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

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

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

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:

$\displaystyle \frac{d (yÂ³)}{dy} \times \frac{dy}{dx}$

(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,

$\displaystyle 3y^Â² \times \frac{dy}{dx} = 1$