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© Matthew Pinkney 2003 |
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MathsRevision.net
Pure Section
Algebra
Calculus
Trigonometry
Geometry
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Implicit DifferentiationIf y³ = x, how would you differentiate this with respect to
x? There are three ways: Method 1Rewrite it as y = x(1/3) and differentiate as
normal (in harder cases, this is not possible!) Method 2Find dx/dy: dx = 3y²
So we get: Method 3Differentiate term by term and use the chain
rule:
The left hand side becomes:
(although it is not strictly 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, 3y² × dy = 1 ExampleDifferentiate x² + y² = 3x, with
respect to x.
ExampleDifferentiate ax with respect to x. You might be tempted to write xax-1 as the answer. This is wrong. That would be the answer if we were differentiating with respect to a not x. Put y = ax . Then, taking logarithms of both sides, we get: ln y = ln (ax) So, differentiating implicitly, we get: (1/y) (dy/dx) = lna |