Differentiation
Introduced| English | Meaning |
|---|---|
| differentiation/ˌdɪfəˌrenʃɪˈeɪʃn/ | The process of finding a derivative, which describes a function's local rate of change. |
| derivative/dɪˈrɪvətɪv/ | The instantaneous rate of change of a function with respect to its input. |
| power rule/ˈpaʊə ruːl/ | The differentiation rule stating that the derivative of x raised to n is n multiplied by x raised to (n - 1). |
| second derivative/ˈsekənd dɪˈrɪvətɪv/ | The derivative of a function's first derivative, describing how its gradient changes. |
| chain rule/tʃeɪn ruːl/ | The rule for differentiating a composite function by multiplying the appropriate derivatives. |
| stationary point/ˈsteɪʃənəri pɔɪnt/ | A point on a differentiable curve where its derivative is zero. |
| optimisation/ˌɒptɪmaɪˈzeɪʃn/ | Finding the greatest or least attainable value of an objective while satisfying the stated constraints. |
Measure change at one instant
- A tank is filling, but its width changes with height. How fast does the water level rise now? An average over a whole minute may hide the answer.
- Differentiation 微分 finds an instantaneous rate. For $y=f(x)$, the derivative 导数 $f'(x)$ or $\dfrac{dy}{dx}$ is the gradient of the tangent.
From a chord to a gradient
- On $y=x^3$, take points with $x=2$ and $x=2+h$. For $h\ne0$, their chord gradient is $\dfrac{(2+h)^3-8}{h}=12+6h+h^2$.
- As $h$ approaches zero from either side, the gradient approaches $12$. This explains the tangent gradient $f'(2)=12$; never substitute $h=0$ into the original fraction.
For y=x³, find chord gradients between x=2 and x=2+h for h=0.1,0.01,−0.01. Use their limit to write the tangent at x=2.
The gradient is $12+6h+h^2$: 12.61,12.0601,11.9401. These approach 12 from either side. The point is (2,8), so the tangent is $y-8=12(x-2)$.
The power rule 幂法则
- For rational $n$, $\dfrac d{dx}x^n=nx^{n-1}$ wherever the function is differentiable. Rewrite roots and reciprocals as powers, and differentiate multiples, sums and differences term by term.
- Differentiate again for the second derivative 二阶导数, written $f''(x)$ or $\dfrac{d^2y}{dx^2}$. A constant differentiates to zero; keep the original domain restrictions.
For $x>0$, $f(x)=3\sqrt{x}+2/x$. Then $f'(x)=\dfrac32x^{-1/2}-2x^{-2}$ and $f''(x)=-\dfrac34x^{-3/2}+4x^{-3}$.
For x>0, differentiate $f(x)=4x^{3/2}-6/x+5$ twice. Find f′(4) and f″(4).
$f\prime(x)=6\sqrt{x}+6/x^2$; $f\prime\prime(x)=3/\sqrt{x}-12/x^3$. At 4 these are $99/8$ and $21/16$.
The chain rule 链式法则
- Differentiate the outside, keep the inside unchanged, then multiply by the derivative of the inside. For $y=\sqrt{2x^3+5}$, this gives $\dfrac{dy}{dx}=\dfrac{3x^2}{\sqrt{2x^3+5}}$ where $2x^3+5>0$.
- For $y=(3x+1)^4$, the inner derivative is $3$, so $\dfrac{dy}{dx}=12(3x+1)^3$. Check this extra factor every time.
If y = x³, then dy/dx = 3x². What is the gradient at x = 2?
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dy/dx = 3x² = 3 × 2² = 3 × 4 = 12.
If y = (2x + 1)³, then dy/dx = 6(2x + 1)². What is dy/dx at x = 1?
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At x = 1: dy/dx = 6(2(1) + 1)² = 6(3)² = 6 × 9 = 54.
Differentiate $y=\sqrt{3x^3+4}$ and find its gradient at x=1. Separately differentiate $(5-2x)^{-2}$.
The first derivative is $9x^2/(2\sqrt{3x^3+4})$, giving $9/(2\sqrt7)$ at 1. The second derivative requested (of the separate expression, not y″) is $4(5-2x)^{-3}$.
Tangents and normals
- Substitute the point's $x$ into the derivative to get $m$. The tangent through $(p,q)$ is $y-q=m(x-p)$. A normal is perpendicular: its gradient is $-1/m$ when $m\ne0$.
- On $y=x^2$ at $(1,1)$, the tangent is $y=2x-1$ and the normal is $y=-x/2+3/2$. If the tangent is horizontal, its normal is the vertical line $x=p$; do not divide by zero.
The gradient at a point
gradient = dy/dx = f ′(x)
Slide the point along the curve — the tangent's slope IS the gradient f ′(x) there.
Find the tangent and normal to $y=x^3-2x+1$ at x=2. Separately find both lines for $y=x^2-6x+4$ at x=3.
The first point is (2,5) and gradient 10. Tangent $y=10x-15$; normal $x+10y=52$. The second point is (3,−5), gradient 0. Tangent $y=-5$ and vertical normal $x=3$.
Stationary points 驻点 and a sketch
- Solve $f'(x)=0$, then substitute each solution into $f(x)$ to find both coordinates. A negative second derivative gives a local maximum; a positive one gives a local minimum.
- For $f=x^3-6x^2+9x+2$, $f'=3(x-1)(x-3)$ and $f''=6x-12$. Thus $(1,6)$ is a maximum and $(3,2)$ a minimum. The derivative is positive outside $(1,3)$ and negative inside it: the curve rises, falls, then rises.
If $f''=0$, this test is inconclusive. For $f=x^4$, $f'(0)=f''(0)=0$, but $f'=4x^3$ changes from negative to positive across zero, proving a minimum at $(0,0)$. For a sketch also mark intercepts and the domain.
The curve y = x² − 6x + 5 has a stationary point. At what x value?
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dy/dx = 2x − 6 = 0 gives x = 3.
At a stationary point, if the second derivative f″(x) > 0, the point is a:
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f″ > 0 means the curve bends upward — a minimum; f″ < 0 gives a maximum.
If dy/dx > 0 for all x in an interval, the function is increasing on that interval.
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A positive derivative means the function is rising — it is increasing.
For y = x³ − 6x² + 9x + 2, the second derivative at x = 3 is:
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d²y/dx² = 6x − 12. At x = 3: 6(3) − 12 = 18 − 12 = 6 > 0, so it's a minimum.
For $f(x)=x^3-12x+16=(x-2)^2(x+4)$, find and classify stationary points, state increasing/decreasing intervals, and sketch with intercepts.
$f\prime=3(x-2)(x+2)$ and $f\prime\prime=6x$. Maximum (−2,32); minimum (2,0). Increasing for x<−2 or x>2, decreasing for −2<x<2. Crosses at (−4,0), touches at (2,0), and crosses the y-axis at (0,16).
Explain why the second-derivative test does not classify the stationary point of y=x⁴ at zero. Classify it by another allowed method.
$y\prime=4x^3$ and $y\prime\prime=12x^2$ are both zero at x=0, so the second-derivative test is inconclusive. For x<0 the first derivative is negative and for x>0 positive; the curve changes from falling to rising, proving a minimum at (0,0).
Connected rates of change
- Write the relation between quantities before using time. If $A=\pi r^2$, then $\dfrac{dA}{dt}=\dfrac{dA}{dr}\dfrac{dr}{dt}=2\pi r\dfrac{dr}{dt}$.
- When $r=5\text{ cm}$ and $\dfrac{dr}{dt}=0.3\,\dfrac{\text{cm}}{\text{s}}$, $\dfrac{dA}{dt}=3\pi\,\dfrac{\text{cm}^2}{\text{s}}$. A shrinking radius has a negative rate; do not lose that sign.
A sphere has $V=4\pi r^3/3$ and $A=4\pi r^2$. Its volume decreases at $12\pi$ cm³/s. At r=2 cm, find dr/dt and dA/dt, with signs and units.
$dV/dt=4\pi r^2\,dr/dt=-12\pi$. At r=2, $dr/dt=-3/4$ cm/s. Then $dA/dt=8\pi r\,dr/dt=-12\pi$ cm²/s.
On $y=ax^{3/2}-8x$ (x>0), x increases at 3 units/s. When x=4, y decreases at 6 units/s. Find a and the x-coordinate of the stationary point.
$dy/dt=(3a\sqrt{x}/2-8)dx/dt$. At x=4, $-6=3(3a-8)$, giving a=2. Then $dy/dx=3\sqrt{x}-8=0$ gives $x=64/9$.
Optimisation 最优化 with a constraint
- A farmer has $60\text{ m}$ of fence for three sides of a rectangle beside a wall. Let each short side be $x$ and the long side $y$: $2x+y=60$, so $A=x(60-2x)$, with $0<x<30$.
- Set $A'=60-4x=0$: $x=15,y=30$. Since $A''=-4<0$, this is a maximum, $450\text{ m}^2$. Check the allowed domain and its boundaries as well as the stationary point.
A closed cylinder has volume 54π cm³. Its total surface area is $S=2\pi r^2+2\pi rh$. Find the dimensions giving minimum surface area and justify the nature of the result.
The volume constraint gives $h=54/r^2$ for r>0. Thus $S=2\pi r^2+108\pi/r$, $S\prime=4\pi r-108\pi/r^2$. The only positive stationary radius is r=3, giving h=6 and S=54π. $S\prime\prime=4\pi+216\pi/r^3>0$ on r>0, so this is the minimum; S grows at both ends of the physical domain.
Check the requested quantity
- A derivative gives a gradient or rate; a stationary point needs coordinates and a nature test. An increasing interval is found from the sign of the first derivative, not the sign of $y$.
- For connected rates, include time and units. For optimisation, first use the constraint to build a function of one variable, then test your answer in its physical domain.
For y = x cubed minus 12x, what is the gradient at x = 1, and what is y at the positive stationary point?
3 minus 12 is -9. At x = 2, 8 minus 24 is -16.
For y=x³−12x, what is y at its positive stationary point?
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The derivative is zero at x = 2 and the value is -16.