链式法则是一种求导的重要方法，需要掌握牢固。
链式法则是一种对组合函数求导的方式，但是通常对于一般的函数，也可以把它转换为组合函数的形式来进行求导。它的具体定义：
If f(u) is differentiable at the point u = g(x) and g(x) is differentiable at x, then the composite function ( f ∘ g ) ( x ) = f ( g ( x ) ) (f \circ g)(x) = f(g(x)) ( f ∘ g ) ( x ) = f ( g ( x ))
( f ∘ g ) ′ ( x ) = f ′ ( g ( x ) ) ⋅ g ′ ( x ) (f \circ g)'(x) = f'(g(x))\cdot g'(x) ( f ∘ g ) ′ ( x ) = f ′ ( g ( x )) ⋅ g ′ ( x )
In Leibniz's notation, if y = f(u) and u = g(x), then
d y d x = d y d u ⋅ d u d x \frac{dy}{dx} = \frac{dy}{du}\cdot \frac{du}{dx} d x d y = d u d y ⋅ d x d u
注意法则的前提，是f(u)和g(x)均为可微的。
例子：求x ( t ) = c o s ( t 2 + 1 ) x(t) = cos(t^2 + 1) x ( t ) = cos ( t 2 + 1 ) t ≥ 0 t \geq 0 t ≥ 0
设u = t 2 + 1 u = t^2 + 1 u = t 2 + 1
d x d t = d x d u ⋅ d u d t = − sin ( u ) ⋅ 2 t = − sin ( t 2 + 1 ) ⋅ 2 t = − 2 t sin ( t 2 + 1 ) \begin{aligned}
\frac{dx}{dt} &= \frac{dx}{du}\cdot \frac{du}{dt} \\
&= -\sin(u)\cdot 2t \\
&= -\sin(t^2 + 1)\cdot 2t \\
&= -2t\sin(t^2 + 1)
\end{aligned} d t d x = d u d x ⋅ d t d u = − sin ( u ) ⋅ 2 t = − sin ( t 2 + 1 ) ⋅ 2 t = − 2 t sin ( t 2 + 1 ) 更进一步的，链式法则可以多次使用，比如：
g ( t ) = t a n ( 5 − sin 2 t ) g(t) = tan(5 - \sin 2t) g ( t ) = t an ( 5 − sin 2 t )
可以先设u = 5 − sin 2 t u = 5 - \sin 2t u = 5 − sin 2 t sin 2 t \sin 2t sin 2 t u = 2 t u = 2t u = 2 t
链式法则也可以应用于指数函数上，称为（Power chain rule），比如：
d d x ( 5 x 3 − x 4 ) 7 = 7 ( 5 x 3 − x 4 ) 6 d d x ( 5 x 3 − x 4 ) \frac{d}{dx}(5x^3 - x^4)^7 = 7(5x^3 - x^4)^6\frac{d}{dx}(5x^3 - x^4) d x d ( 5 x 3 − x 4 ) 7 = 7 ( 5 x 3 − x 4 ) 6 d x d ( 5 x 3 − x 4 )
参数方程（parametric equations）是另一种表述数量关系的方式，比如知道x和y分别与另一个变量t的关系，但是还不清楚x与y之间的关系，那么就有函数：
x = f ( t ) , y = g ( t ) x = f(t), y = g(t) x = f ( t ) , y = g ( t )
这称为参数方程。如果我们知道f和g在t处可微，也知道y在x处可微 ，那么三者之间的关系为：
d y d t = d y d x ⋅ d x d t \frac{dy}{dt} = \frac{dy}{dx}\cdot \frac{dx}{dt} d t d y = d x d y ⋅ d t d x
或者说：
d y d x = d y / d t d x / d t \frac{dy}{dx} = \frac{dy/dt}{dx/dt} d x d y = d x / d t d y / d t
这个定理的好处在于，我们不需要知道y=f(x)，就可以求得d y / d x dy/dx d y / d x
更近一步的，二阶微分函数可以通过如下定理求得：
d 2 y d x 2 = d y ′ / d t d x / d t \frac{d^2y}{dx^2} = \frac{dy'/dt}{dx/dt} d x 2 d 2 y = d x / d t d y ′ / d t
注意这里：
y ′ = d y d x y' = \frac{dy}{dx} y ′ = d x d y
例子：求x = t − t 2 , y = t − t 3 x = t - t^2, y = t - t^3 x = t − t 2 , y = t − t 3
y ′ = d y d x = d y / d t d x / d t = 1 − 3 t 2 1 − 2 t y' = \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1 - 3t^2}{1 - 2t} y ′ = d x d y = d x / d t d y / d t = 1 − 2 t 1 − 3 t 2
d y ′ d t = d d t ( 1 − 3 t 2 1 − 2 t ) = 2 − 6 t + 6 t 2 ( 1 − 2 t ) 2 \frac{dy'}{dt} = \frac{d}{dt}(\frac{1 - 3t^2}{1 - 2t}) = \frac{2 - 6t + 6t^2}{(1 - 2t)^2} d t d y ′ = d t d ( 1 − 2 t 1 − 3 t 2 ) = ( 1 − 2 t ) 2 2 − 6 t + 6 t 2
d 2 y d x 2 = d y ′ / d t d x / d t = ( 2 − 6 t + 6 t 2 ) / ( 1 − 2 t ) 2 1 − 2 t = 2 − 6 t + 6 t 2 ( 1 − 2 t ) 3 \frac{d^2y}{dx^2} = \frac{dy'/dt}{dx/dt} = \frac{(2 - 6t + 6t^2)/(1 - 2t)^2}{1 - 2t} = \frac{2 - 6t + 6t^2}{(1 - 2t)^3} d x 2 d 2 y = d x / d t d y ′ / d t = 1 − 2 t ( 2 − 6 t + 6 t 2 ) / ( 1 − 2 t ) 2 = ( 1 − 2 t ) 3 2 − 6 t + 6 t 2
练习
y = 6 u − 9 , u = ( 1 / 2 ) x 4 y = 6u - 9, u = (1/2)x^4 y = 6 u − 9 , u = ( 1/2 ) x 4
d y d x = d y d u ⋅ d u d x = 6 ⋅ 2 x 3 = 12 x 3 \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 6 \cdot 2x^3 = 12x^3 d x d y = d u d y ⋅ d x d u = 6 ⋅ 2 x 3 = 12 x 3
y = 2 u 3 , u = 8 x − 1 y = 2u^3, u = 8x - 1 y = 2 u 3 , u = 8 x − 1
d y d x = d y d u ⋅ d u d x = 6 u 2 ⋅ 8 = 48 ( 8 x − 1 ) 2 \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 6u^2 \cdot 8 = 48(8x - 1)^2 d x d y = d u d y ⋅ d x d u = 6 u 2 ⋅ 8 = 48 ( 8 x − 1 ) 2
y = sin u , u = 3 x + 1 y = \sin u, u = 3x + 1 y = sin u , u = 3 x + 1
d y d x = d y d u ⋅ d u d x = cos u ⋅ 3 = 3 cos ( 3 x + 1 ) \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos u \cdot 3 = 3\cos(3x + 1) d x d y = d u d y ⋅ d x d u = cos u ⋅ 3 = 3 cos ( 3 x + 1 )
y = cos u , u = − x / 3 y = \cos u, u = -x/3 y = cos u , u = − x /3
d y d x = d y d u ⋅ d u d x = − sin u ⋅ − 1 3 = 1 3 sin ( − x 3 ) \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\sin u \cdot -\frac{1}{3} = \frac{1}{3}\sin(-\frac{x}{3}) d x d y = d u d y ⋅ d x d u = − sin u ⋅ − 3 1 = 3 1 sin ( − 3 x )
y = cos u , u = sin x y = \cos u, u = \sin x y = cos u , u = sin x
d y d x = d y d u ⋅ d u d x = − sin u ⋅ cos x = − sin ( sin x ) ⋅ cos x \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\sin u \cdot \cos x = -\sin(\sin x)\cdot \cos x d x d y = d u d y ⋅ d x d u = − sin u ⋅ cos x = − sin ( sin x ) ⋅ cos x
y = sin u , u = x − cos x y = \sin u, u = x - \cos x y = sin u , u = x − cos x
d y d x = cos u ⋅ ( 1 + sin x ) = cos ( x − cos x ) ⋅ ( 1 + sin x ) \frac{dy}{dx} = \cos u \cdot (1 + \sin x) = \cos (x - \cos x) \cdot (1 + \sin x) d x d y = cos u ⋅ ( 1 + sin x ) = cos ( x − cos x ) ⋅ ( 1 + sin x )
y = tan u , u = 10 x − 5 y = \tan u, u = 10x - 5 y = tan u , u = 10 x − 5
d y d x = sec 2 u ⋅ 10 = 10 sec 2 ( 10 x − 5 ) \frac{dy}{dx} = \sec^2u \cdot 10 = 10\sec^2(10x - 5) d x d y = sec 2 u ⋅ 10 = 10 sec 2 ( 10 x − 5 )
y = − sec u , u = x 2 + 7 x y = -\sec u, u = x^2 + 7x y = − sec u , u = x 2 + 7 x
d y d x = − sec u tan u ⋅ ( 2 x + 7 ) = − sec ( x 2 + 7 x ) tan ( x 2 + 7 x ) ⋅ ( 2 x + 7 ) \frac{dy}{dx} = -\sec u \tan u \cdot (2x + 7) = -\sec(x^2 + 7x)\tan(x^2 + 7x)\cdot (2x + 7) d x d y = − sec u tan u ⋅ ( 2 x + 7 ) = − sec ( x 2 + 7 x ) tan ( x 2 + 7 x ) ⋅ ( 2 x + 7 )
y = ( 2 x + 1 ) 5 y = (2x + 1)^5 y = ( 2 x + 1 ) 5
y ′ = 5 ( 2 x + 1 ) 4 ⋅ 2 = 10 ( 2 x + 1 ) 4 y' = 5(2x + 1)^4 \cdot 2 = 10(2x + 1)^4 y ′ = 5 ( 2 x + 1 ) 4 ⋅ 2 = 10 ( 2 x + 1 ) 4
y = ( 4 − 3 x ) 9 y = (4 - 3x)^9 y = ( 4 − 3 x ) 9
y ′ = − 27 ( 4 − 3 x ) 8 y' = -27(4 - 3x)^8 y ′ = − 27 ( 4 − 3 x ) 8
y = ( 1 − x 7 ) − 7 y = (1 - \frac{x}{7})^{-7} y = ( 1 − 7 x ) − 7
y ′ = − 7 ( 1 − x 7 ) − 8 ⋅ − 1 7 = x 7 ) − 8 y' = -7(1 - \frac{x}{7})^{-8}\cdot -\frac{1}{7} = \frac{x}{7})^{-8} y ′ = − 7 ( 1 − 7 x ) − 8 ⋅ − 7 1 = 7 x ) − 8
y = ( x 2 − 1 ) − 10 y = (\frac{x}{2} - 1)^{-10} y = ( 2 x − 1 ) − 10
y ′ = − 10 ( x 2 − 1 ) − 11 ⋅ 1 2 = − 5 ( x 2 − 1 ) − 11 y' = -10(\frac{x}{2} - 1)^{-11}\cdot \frac{1}{2} = -5(\frac{x}{2} - 1)^{-11} y ′ = − 10 ( 2 x − 1 ) − 11 ⋅ 2 1 = − 5 ( 2 x − 1 ) − 11
y = ( x 2 8 + x − 1 x ) 4 y = (\frac{x^2}{8} + x - \frac{1}{x})^4 y = ( 8 x 2 + x − x 1 ) 4
y ′ = 4 ( x 2 8 + x − 1 x ) 3 ⋅ ( 4 x + 1 + 1 x 2 ) y' = 4(\frac{x^2}{8} + x - \frac{1}{x})^3 \cdot (4x + 1 + \frac{1}{x^2}) y ′ = 4 ( 8 x 2 + x − x 1 ) 3 ⋅ ( 4 x + 1 + x 2 1 )
y = ( x 5 + 1 5 x ) 5 y = (\frac{x}{5} + \frac{1}{5x})^5 y = ( 5 x + 5 x 1 ) 5
y ′ = 5 ( x 5 + 1 5 x ) 4 ⋅ ( 1 5 − 1 5 x 2 ) y' = 5(\frac{x}{5} + \frac{1}{5x})^4 \cdot (\frac{1}{5} - \frac{1}{5x^2}) y ′ = 5 ( 5 x + 5 x 1 ) 4 ⋅ ( 5 1 − 5 x 2 1 )
y = sec ( tan x ) y = \sec(\tan x) y = sec ( tan x )
y ′ = sec ( tan x ) tan ( tan x ) ⋅ sec 2 x y' = \sec(\tan x)\tan(\tan x) \cdot \sec^2 x y ′ = sec ( tan x ) tan ( tan x ) ⋅ sec 2 x
y = cot ( π − 1 x ) y = \cot (\pi - \frac{1}{x}) y = cot ( π − x 1 )
y ′ = − csc 2 ( π − 1 x ) ⋅ 1 x 2 y' = -\csc^2(\pi - \frac{1}{x})\cdot \frac{1}{x^2} y ′ = − csc 2 ( π − x 1 ) ⋅ x 2 1
y = sin 3 x y = \sin^3 x y = sin 3 x
y ′ = 3 sin 2 x ⋅ cos x y' = 3\sin^2 x \cdot \cos x y ′ = 3 sin 2 x ⋅ cos x
y = 5 cos − 4 x y = 5\cos^{-4}x y = 5 cos − 4 x
y ′ = − 20 cos − 5 x ⋅ − sin x = 20 cos − 5 x ⋅ sin x y' = -20\cos^{-5}x \cdot -\sin x = 20\cos^{-5}x\cdot \sin x y ′ = − 20 cos − 5 x ⋅ − sin x = 20 cos − 5 x ⋅ sin x
p = 3 − t p = \sqrt{3 - t} p = 3 − t
p ′ = − 1 2 3 − t p' = -\frac{1}{2\sqrt{3 - t}} p ′ = − 2 3 − t 1
q = 2 r − r 2 q = \sqrt{2r - r^2} q = 2 r − r 2
q ′ = 1 2 2 r − r 2 ⋅ ( 2 − 2 r ) q' = \frac{1}{2\sqrt{2r - r^2}} \cdot (2 - 2r) q ′ = 2 2 r − r 2 1 ⋅ ( 2 − 2 r )
s = 4 3 π sin 3 t + 4 5 π cos 5 t s = \frac{4}{3\pi}\sin 3t + \frac{4}{5\pi}\cos 5t s = 3 π 4 sin 3 t + 5 π 4 cos 5 t
s ′ = 4 π cos 3 t − 4 π sin 5 t s' = \frac{4}{\pi}\cos 3t - \frac{4}{\pi}\sin 5t s ′ = π 4 cos 3 t − π 4 sin 5 t
s = sin ( 3 π t 2 ) + cos ( 3 π t 2 ) s = \sin(\frac{3\pi t}{2}) + \cos(\frac{3\pi t}{2}) s = sin ( 2 3 π t ) + cos ( 2 3 π t )
s ′ = 3 π 2 cos ( 3 π t 2 ) − 3 π 2 sin ( 3 π t 2 ) s' = \frac{3\pi}{2}\cos(\frac{3\pi t}{2}) - \frac{3\pi}{2}\sin(\frac{3\pi t}{2}) s ′ = 2 3 π cos ( 2 3 π t ) − 2 3 π sin ( 2 3 π t )
23 - 66 略
x = cos 2 t , y = sin 2 t , 0 ≤ t ≤ π x = \cos 2t, y = \sin 2t, 0 \leq t \leq \pi x = cos 2 t , y = sin 2 t , 0 ≤ t ≤ π
∵ x 2 + y 2 = cos 2 2 t + sin 2 2 t = 1 , and 0 ≤ 2 t ≤ 2 π ∴ x 2 + y 2 = 1 , is a full circle. \because
x^2 + y^2 = \cos^2 2t + \sin^2 2t = 1, \text{and}\\
0 \leq 2t \leq 2\pi \\
\therefore x^2 + y^2 = 1, \text{is a full circle.} ∵ x 2 + y 2 = cos 2 2 t + sin 2 2 t = 1 , and 0 ≤ 2 t ≤ 2 π ∴ x 2 + y 2 = 1 , is a full circle. 质点从(1, 0)处开始，逆时针绕圆一周。
x = cos ( π − t ) , y = sin ( π − 1 ) , 0 ≤ t ≤ π x = \cos(\pi - t), y = \sin(\pi - 1), 0 \leq t \leq \pi x = cos ( π − t ) , y = sin ( π − 1 ) , 0 ≤ t ≤ π
x 2 + y 2 = 1 x^2 + y^2 = 1 x 2 + y 2 = 1
质点从(-1, 0)处开始，顺时针绕圆半圈。
x = 4 cos t , y = 2 sin t , 0 ≤ t ≤ 2 π x = 4\cos t, y = 2\sin t, 0 \leq t \leq 2\pi x = 4 cos t , y = 2 sin t , 0 ≤ t ≤ 2 π
( x 2 ) 2 + y 2 = 4 cos 2 t + 4 sin 2 t = 4 (\frac{x}{2})^2 + y^2 = 4\cos^2 t + 4\sin^2 t = 4 ( 2 x ) 2 + y 2 = 4 cos 2 t + 4 sin 2 t = 4
是一个在x轴方向被拉伸的椭圆，质点从(4, 0)处开始，逆时针绕成一个椭圆。
x = 4 sin t , y = 5 cos t , 0 ≤ t ≤ 2 π x = 4\sin t, y = 5\cos t, 0 \leq t \leq 2\pi x = 4 sin t , y = 5 cos t , 0 ≤ t ≤ 2 π
( x 4 ) 2 + ( y 5 ) 2 = 1 (\frac{x}{4})^2 +(\frac{y}{5})^2 = 1 ( 4 x ) 2 + ( 5 y ) 2 = 1
质点从(0, 5)开始，顺时针旋转一周。
x = 3 t , y = 9 t 2 , − ∞ < t < + ∞ x = 3t, y = 9t^2, -\infty < t < +\infty x = 3 t , y = 9 t 2 , − ∞ < t < + ∞
y = 9 ( x 3 ) 2 = x 2 y = 9(\frac{x}{3})^2 = x^2 y = 9 ( 3 x ) 2 = x 2
x = − t , y = t , t ≥ 0 x = -\sqrt{t}, y = t, t \geq 0 x = − t , y = t , t ≥ 0
y = x 2 , x ≤ 0 y = x^2, x \leq 0 y = x 2 , x ≤ 0
x = 2 t − 5 , y = 4 t − 7 , − ∞ < t < + ∞ x = 2t - 5, y = 4t - 7, -\infty < t < +\infty x = 2 t − 5 , y = 4 t − 7 , − ∞ < t < + ∞
y − 2 x = 3 , y = 2 x + 3 y - 2x = 3, y = 2x + 3 y − 2 x = 3 , y = 2 x + 3
x = 3 − 3 t , y = 2 t , 0 ≤ t ≤ 1 x = 3 - 3t, y = 2t, 0 \leq t \leq 1 x = 3 − 3 t , y = 2 t , 0 ≤ t ≤ 1
y = 2 − 2 3 x y = 2 - \frac{2}{3}x y = 2 − 3 2 x
75 - 86 略
x = 2 cos t , y = 2 sin t , t = π / 4 x = 2\cos t, y = 2\sin t, t = \pi/4 x = 2 cos t , y = 2 sin t , t = π /4
d y d x = d y / d t d x / d t = 2 cos t − 2 sin t \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\cos t}{-2\sin t} d x d y = d x / d t d y / d t = − 2 sin t 2 cos t 代入t = π / 4 t = \pi/4 t = π /4 ( 2 , 2 ) (\sqrt{2}, \sqrt{2}) ( 2 , 2 )
− 1 = y − 2 x − 2 y = 2 2 − x -1 = \frac{y - \sqrt{2}}{x - \sqrt{2}} \\
y = 2\sqrt{2} - x − 1 = x − 2 y − 2 y = 2 2 − x d 2 y d x 2 = d y ′ / d t d x / d t = 1 / sin 2 t − 2 sin t = − 2 sin 3 t \frac{d^2y}{dx^2} = \frac{dy'/dt}{dx/dt} = \frac{1/\sin^2 t}{-2\sin t} = -\frac{2}{\sin^3 t} d x 2 d 2 y = d x / d t d y ′ / d t = − 2 sin t 1/ sin 2 t = − sin 3 t 2
x = cos t , y = 3 cos t , t = 2 π / 3 x = \cos t, y = \sqrt{3}\cos t, t = 2\pi/3 x = cos t , y = 3 cos t , t = 2 π /3
d y d x = d y / d t d x / d t = − 3 sin t − sin t = 3 \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-\sqrt{3}\sin t}{-\sin t} = \sqrt{3} d x d y = d x / d t d y / d t = − sin t − 3 sin t = 3 斜率为3 \sqrt{3} 3 t = 2 π / 3 t = 2\pi/3 t = 2 π /3 ( − 1 / 2 , − 3 / 2 ) (-1/2, -\sqrt{3}/2) ( − 1/2 , − 3 /2 )
3 = y + 3 / 2 x + 1 / 2 y = 3 x \sqrt{3} = \frac{y + \sqrt{3}/2}{x + 1/2} \\
y = \sqrt{3}x 3 = x + 1/2 y + 3 /2 y = 3 x d 2 y d x 2 = d y ′ / d t d x / d t = 0 \frac{d^2y}{dx^2} = \frac{dy'/dt}{dx/dt} = 0 d x 2 d 2 y = d x / d t d y ′ / d t = 0
x = t , y = t , t = 1 / 4 x = t, y = \sqrt{t}, t = 1/4 x = t , y = t , t = 1/4
d y d x = d y / d t d x / d t = 1 2 t 1 = 1 2 t \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{1}{2\sqrt{t}}}{1} = \frac{1}{2\sqrt{t}} d x d y = d x / d t d y / d t = 1 2 t 1 = 2 t 1 代入t = 1 / 4 t = 1/4 t = 1/4 ( 1 / 4 , 1 / 2 ) (1/4, 1/2) ( 1/4 , 1/2 )
1 4 = y − 1 / 2 x − 1 / 4 y = 1 4 x − 7 16 \frac{1}{4} = \frac{y - 1/2}{x - 1/4} \\
y = \frac{1}{4}x - \frac{7}{16} 4 1 = x − 1/4 y − 1/2 y = 4 1 x − 16 7 d 2 y d x 2 = d y ′ / d t d x / d t = − 1 4 x − 3 2 \frac{d^2y}{dx^2} = \frac{dy'/dt}{dx/dt} = -\frac{1}{4}x^{-\frac{3}{2}} d x 2 d 2 y = d x / d t d y ′ / d t = − 4 1 x − 2 3
x = − t + 1 , y = 3 t , t = 3 x = -\sqrt{t + 1}, y = \sqrt{3t}, t = 3 x = − t + 1 , y = 3 t , t = 3
d y d x = d y / d t d x / d t = 1 2 ( 3 t ) − 1 / 2 − 1 2 ( t + 1 ) − 1 / 2 = − ( 3 t t + 1 ) − 1 / 2 \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{1}{2}(3t)^{-1/2}}{-\frac{1}{2}(t + 1)^{-1/2}} = -(\frac{3t}{t + 1})^{-1/2} d x d y = d x / d t d y / d t = − 2 1 ( t + 1 ) − 1/2 2 1 ( 3 t ) − 1/2 = − ( t + 1 3 t ) − 1/2 d 2 y d x 2 = d y ′ / d t d x / d t = 1 2 ( 3 t t + 1 ) − 3 / 2 − 1 2 ( t + 1 ) − 1 / 2 = − ( 3 t t + 1 ) − 3 / 2 ( t + 1 ) − 1 / 2 \frac{d^2y}{dx^2} = \frac{dy'/dt}{dx/dt} = \frac{\frac{1}{2}(\frac{3t}{t + 1})^{-3/2}}{-\frac{1}{2}(t + 1)^{-1/2}} = -\frac{(\frac{3t}{t + 1})^{-3/2}}{(t + 1)^{-1/2}} d x 2 d 2 y = d x / d t d y ′ / d t = − 2 1 ( t + 1 ) − 1/2 2 1 ( t + 1 3 t ) − 3/2 = − ( t + 1 ) − 1/2 ( t + 1 3 t ) − 3/2
x = 2 t 2 + 3 , y = t 4 , t = − 1 x = 2t^2 + 3, y = t^4, t = -1 x = 2 t 2 + 3 , y = t 4 , t = − 1
d y d x = d y / d t d x / d t = 2 t 3 3 t = 2 3 t 2 \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t^3}{3t} = \frac{2}{3}t^2 d x d y = d x / d t d y / d t = 3 t 2 t 3 = 3 2 t 2 d 2 y d x 2 = d y ′ / d t d x / d t = 4 / 3 t 4 t = 1 / 3 \frac{d^2y}{dx^2} = \frac{dy'/dt}{dx/dt} = \frac{4/3t}{4t} = 1/3 d x 2 d 2 y = d x / d t d y ′ / d t = 4 t 4/3 t = 1/3
x = t − sin x , y = 1 − cos t , t = π / 3 x = t - \sin x, y = 1 - \cos t, t = \pi/3 x = t − sin x , y = 1 − cos t , t = π /3
d y d x = d y / d t d x / d t = − sin t cos t \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = -\frac{\sin t}{\cos t} d x d y = d x / d t d y / d t = − cos t sin t d 2 y d x 2 = d y ′ / d t d x / d t = − cos 2 t + sin 2 t cos 2 t − cos t = 1 cos 3 t \frac{d^2y}{dx^2} = \frac{dy'/dt}{dx/dt} = \frac{-\frac{\cos^2 t + \sin^2 t}{\cos^2 t}}{-\cos t} = \frac{1}{\cos^3 t} d x 2 d 2 y = d x / d t d y ′ / d t = − cos t − c o s 2 t c o s 2 t + s i n 2 t = cos 3 t 1
x = cos t , y = 1 + sin t , t = π / 2 x = \cos t, y = 1 + \sin t, t = \pi/2 x = cos t , y = 1 + sin t , t = π /2
d y d x = d y / d t d x / d t = cos t − sin t \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\cos t}{-\sin t} d x d y = d x / d t d y / d t = − sin t cos t d 2 y d x 2 = d y ′ / d t d x / d t = 1 sin 3 t \frac{d^2y}{dx^2} = \frac{dy'/dt}{dx/dt} = \frac{1}{\sin^3 t} d x 2 d 2 y = d x / d t d y ′ / d t = sin 3 t 1
x = sec 2 − 1 , y = tan t , t = − π / 4 x = \sec^2 - 1, y = \tan t, t = -\pi/4 x = sec 2 − 1 , y = tan t , t = − π /4
d y d x = d y / d t d x / d t = sec 2 t 2 sec 2 t tan t = 1 2 tan t \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sec^2 t}{2\sec^2 t\tan t} = \frac{1}{2\tan t} d x d y = d x / d t d y / d t = 2 sec 2 t tan t sec 2 t = 2 tan t 1 d 2 y d x 2 = d y ′ / d t d x / d t = − 1 2 tan − 2 t sec 2 t 2 sec 2 t tan t = 1 4 tan 3 t \frac{d^2y}{dx^2} = \frac{dy'/dt}{dx/dt} = \frac{-\frac{1}{2}\tan^{-2}t\sec^2 t}{2\sec^2 t\tan t} = \frac{1}{4\tan^3 t} d x 2 d 2 y = d x / d t d y ′ / d t = 2 sec 2 t tan t − 2 1 tan − 2 t sec 2 t = 4 tan 3 t 1 
