Differentiation (Pure 3)
| English | Chinese | Pinyin |
|---|---|---|
| derivative | 导数 | dǎo shù |
| inverse tangent | 反正切 | fǎn zhèng qiè |
| chain rule | 链式法则 | liàn shì fǎ zé |
| parametric | 参数 | cān shù |
| product rule | 乘积法则 | chéng jī fǎ zé |
| reciprocal | 倒数 | dào shǔ |
| implicit differentiation | 隐函数微分 | yǐn hán shù wēi fēn |
The derivative 导数 that completes the circle
- You know the derivatives of $\sin$, $\cos$, $\tan$, $e^x$, and $\ln x$. Pure 3 adds one more: the inverse tangent 反正切.
- Together, these standard derivatives let you differentiate any combination of elementary functions.
Review of Pure 2 methods
- The methods are those of Pure 2: the product, quotient and chain rules 链式法则, with parametric 参数 and implicit curves.
- These are the workhorses of differentiation — master them and you can handle any problem.
Worked example. $y = x^2 e^{3x}$. Product rule 乘积法则 with $u = x^2$, $v = e^{3x}$: $\dfrac{dy}{dx} = 2x \cdot e^{3x} + x^2 \cdot 3e^{3x} = e^{3x}(2x + 3x^2)$.
z on an Argand diagram: modulus is the distance, argument the angle, conjugate the reflection
The gradient at a point
gradient = dy/dx
Implicit or not, the derivative is still the slope of the tangent — slide the point to see it.
Pure 3 differentiation still uses the product, quotient and chain rules from Pure 2.
Those rules carry over; only the inverse-tangent derivative is new.
If y = x²e^(3x), then dy/dx = e^(3x)(2x + 3x²). At x = 0, dy/dx = ?
At x = 0: e⁰(0 + 0) = 1 × 0 = 0.
The inverse tangent derivative
- One new standard derivative:
- This appears in integration (the reverse) and in related rates problems.
$\tan^{-1}x$ is NOT $\dfrac{1}{\tan x}$. The notation $\tan^{-1}x$ means the inverse function (arctan), not the reciprocal 倒数. The reciprocal is $\cot x$.
The derivative of tan⁻¹x is 1/(1+x²). What is its value at x = 1?
1/(1 + 1²) = 1/2 = 0.5.
What is the derivative of tan⁻¹x at x = 0?
1/(1 + 0²) = 1/1 = 1.
tan⁻¹x means 1/tan x.
tan⁻¹x is the inverse function (arctan), not the reciprocal. The reciprocal is cot x.
Implicit differentiation 隐函数微分 revisited
- For equations like $x^2 + y^2 = 25$, differentiate every term with respect to $x$.
- When you meet a $y$ term, apply the chain rule: $\dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}$.
- Then solve for $\dfrac{dy}{dx}$.
allow_no_figure: Review lesson consolidating Pure 2 differentiation methods (product, quotient, chain rules) with one new derivative — no new visual concepts introduced.
Worked example — implicit differentiation
- Find $\dfrac{dy}{dx}$ for $x^2 + xy + y^2 = 7$.
- Differentiate: $2x + y + x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0$.
- Collect $\dfrac{dy}{dx}$ terms: $(x + 2y)\dfrac{dy}{dx} = -(2x + y)$.
- $\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$.
- Differentiate relations given implicitly or parametrically.
When differentiating y² implicitly with respect to x, you get:
By the chain rule: d/dx(y²) = 2y · dy/dx.
You've got it
- reuse the product / quotient / chain rules from Pure 2
- the new derivative: $\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1 + x^2}$
- still handle parametric and implicit curves the same way