Flashcards · 8 cards

Differentiation

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Rule
Differentiating from first principles?
Answer
$\dfrac{dy}{dx} = \lim\limits_{h \to 0}\dfrac{f(x + h) - f(x)}{h}$ . Simplify until you can divide by $h$ , then let $h \to 0$ .
f′(x) = lim_h→0 [f(x + h) − f(x)] ⁄ h
1. find f(x + h)2. take away f(x)3. divide by h4. let h → 0
From first principlesSimplify until you can divide by h, then let h → 0
From the lesson: Differentiation from first principles
Rule
Why does that limit give the gradient?
Answer
It is the gradient of the chord from $x$ to $x + h$ . As $h \to 0$ , the chord becomes the tangent.
Chord to tangentAs Q slides towards P (h → 0), the chord's gradient becomes the tangent's
From the lesson: Differentiation from first principles
Rule
Find $\lim\limits_{x \to 3}\dfrac{x^2 - 9}{x - 3}$ .
Answer
Putting in 3 gives $\frac00$ : that means factorise and cancel first. $\dfrac{(x - 3)(x + 3)}{x - 3} = x + 3$ , which tends to 6.
A gap in the graph(x² − 9) ÷ (x − 3) is x + 3 with a gap at x = 3: the limit is the height of the gap, 6
From the lesson: Differentiation from first principles
Rule
Find $\lim\limits_{x \to \infty}\dfrac{2x + 1}{x + 1}$ .
Answer
Divide top and bottom by $x$ : $\dfrac{2 + \frac1x}{1 + \frac1x}$ , which tends to $\dfrac21 = 2$ .
A limit at infinityAs x grows, (2x + 1) ÷ (x + 1) gets as close to 2 as you like
From the lesson: Differentiation from first principles
Rule
Differentiating $y^2$ and $xy$ with respect to $x$ ?
Answer
$y^2$ gives $2y\dfrac{dy}{dx}$ . $xy$ needs the product rule: $y + x\dfrac{dy}{dx}$ .
d/dx (y²) = 2y · dy/dx
d/dx (xy) = y + x · dy/dx
differentiate a y term as usual, then multiply by dy/dx; xy needs the product rule
Terms in yy² gives 2y dy/dx; xy needs the product rule
From the lesson: The rules, implicit differentiation and second derivatives
Rule
The quotient rule?
Answer
For $y = \dfrac uv$ : $\dfrac{dy}{dx} = \dfrac{vu' - uv'}{v^2}$ .
d/dx (u ⁄ v) = (v·u′ − u·v′) ⁄ v²
bottom × derivative of top, minus top × derivative of bottom, all over bottom squared
The quotient ruleBottom × top′ − top × bottom′, over bottom²
From the lesson: The rules, implicit differentiation and second derivatives
Rule
The gradient of the normal at a point?
Answer
$-1 \div$ (gradient of the tangent): the normal is perpendicular to the tangent.
Tangent and normalGradients multiply to −1: m × (−1/m) = −1
From the lesson: The rules, implicit differentiation and second derivatives
Which method?
WAEC 2020 · Paper 2 · Q1
If $x^2 - xy - y^2 = -31$ , find $\dfrac{dy}{dx}$ at $(2, 5)$ .
How do you differentiate the $-xy$ term?
Answer
With the product rule: $-\left(y + x\dfrac{dy}{dx}\right)$ . Collect the $\dfrac{dy}{dx}$ terms on one side, then put in $(2, 5)$ .
From the lesson: The rules, implicit differentiation and second derivatives Try the question