A LevelMathsEdexcelPureFlashcardsDifferentiationImplicit Differentiation

Implicit Differentiation (Edexcel A Level Maths: Pure): Flashcards

Exam code: 9MA0

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FrontImplicit Differentiation
What does it mean to differentiate an equation implicitly?

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What does it mean to differentiate an equation implicitly?
Differentiating both sides with respect to $x$ without first rearranging the equation into the form $y = f (x)$ .
It is used whenever writing $y$ explicitly in terms of $x$ would be awkward or impossible, as for $x^{2} + y^{2} = 25$ .
Complete the rule for differentiating a function of $y$ with respect to $x$ :
$\frac{d}{d x} (f (y)) = f' (y) \times______$
The completed rule is:
$\frac{d}{d x} (f (y)) = f' (y) \frac{d y}{d x}$
This is the chain rule: $y$ is itself a function of $x$ , so differentiating anything built from $y$ brings out a factor of $\frac{d y}{d x}$ .
What is $\frac{d}{d x} (y^{3})$ ?
$3 y^{2} \frac{d y}{d x}$ .
Differentiate the power exactly as usual, then multiply by $\frac{d y}{d x}$ because the variable being differentiated is $y$ rather than $x$ .
What is $\frac{d}{d x} (x y)$ ?
$x \frac{d y}{d x} + y$ .
A term containing both variables is a product, so it needs the product rule as well: differentiating $x$ gives $1$ , and differentiating $y$ gives $\frac{d y}{d x}$ .
Why must you never split $\frac{d y}{d x}$ when rearranging?
Because it is a single algebraic object, not a fraction with $d y$ on top and $d x$ underneath.
Collect it and factorise it out in exactly the way you would treat any single unknown letter.
Will implicit differentiation always give $\frac{d y}{d x}$ as a function of $x$ alone?
No. The answer is usually in terms of both $x$ and $y$ , and that is perfectly acceptable.
To get a numerical gradient you substitute both coordinates of the point, rather than just the $x$ -value.
True or False?
Implicit differentiation is just the chain rule applied to terms containing $y$ .
True.
Every step comes from treating $y$ as a function of $x$ and applying the chain rule, with the product rule joining in for terms that contain both variables.
There is no new rule to learn here, only a new situation in which to use the old ones.

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