Curriculum

Mathematics

Unit 2

0/14

Unit 5

0/12

Text lesson

Implicit Differentiation, Parametric Differentiation and Higher Derivatives

Mathematics

Implicit Differentiation, Parametric Differentiation and Higher Derivatives

Overview

This lesson covers lessons 1-1 and 1-2 of the calculus unit "Differentiation and its Applications".
After it you should be able to:
differentiate a relation in $x$ and $y$ when $y$ cannot be isolated;
find $\frac{d y}{d x}$ for a curve given by a parameter, and the derivative of one function with respect to another;
find second, third and $n$ th derivatives, including for implicit and parametric curves.

🌳 Concept tree

Implicit Differentiation
Parametric Differentiation and Higher Derivatives

├── 1-1 (a) Explicit and implicit relations

│ ├── Explicit functions

│ └── Implicit relations

├── 1-1 (b) Implicit differentiation

│ ├── The method

│ ├── The chain-rule step

│ └── Important remarks

├── 1-1 (c) Parametric differentiation

│ ├── Parametric curves

│ └── The rule

├── 1-1 (d) Derivative of one function with respect to another

│ └── The idea

├── 1-2 (a) Higher-order derivatives

│ ├── Definition and notation

│ ├── A common confusion

│ ├── Patterns for the nth derivative

│ └── Proving relations between derivatives

├── 1-2 (b) Second derivative of implicit relations

│ └── Method

├── 1-2 (c) Second derivative of parametric curves

│ └── The rule

└── 1-2 (d) Meaning of the derivatives

├── Graph connection

└── Summary table

Implicit Differentiation, Parametric Differentiation and Higher Derivatives · 1 / 9

1-1 (a) Explicit and implicit relations

Explicit functions

An explicit function gives $y$ directly in terms of $x$ , for example $y = 4 x^{3} - 5 x + 2$ or $y = \sin (2 x + 3)$ .
For such a function the derivative is found by the ordinary rules.

An implicit relation links $x$ and $y$ in one equation $f (x, y) = 0$ , for example $x y + y - 4 = 0$ or $x^{2} + y^{2} = 9$ .
Sometimes y can be isolated xy+y−4=0 gives y=4x+1 for x≠−1.
Sometimes one equation hides several functions x2+y2=9 is a circle, which fails the vertical-line test, but it splits into two functions.
The upper half is $y = \sqrt{9 - x^{2}}$ (domain $[- 3, 3]$ , range $[0, 3]$ ) and the lower half is $y = - \sqrt{9 - x^{2}}$ (range $[- 3, 0]$ ).
Each half is differentiable for $x \in] - 3, 3 [$ .
In many equations, such as $x^{3} + y^{3} = 6 x y$ , isolating $y$ is impossible or very awkward, so we differentiate the equation as it stands.

Implicit Differentiation, Parametric Differentiation and Higher Derivatives · 2 / 9

1-1 (b) Implicit differentiation

The method

Differentiate both sides of the equation with respect to $x$ .
Treat $y$ as a differentiable function of $x$ and apply the chain rule to every term containing $y$ .
Collect all terms containing $\frac{d y}{d x}$ on one side, factorise, and divide.

Implicit Differentiation, Parametric Differentiation: a circular track and a tangent rod at (3, 4) show implicit and parametric differentiation: the slope is -3/4.

The chain-rule step

$\frac{d}{d x} (y^{n}) = n y^{n - 1} \frac{d y}{d x}$ .
$\frac{d}{d x} (\sin y) = \cos y \frac{d y}{d x}$ and $\frac{d}{d x} (e^{y}) = e^{y} \frac{d y}{d x}$ .
$\frac{d}{d x} (x y) = y + x \frac{d y}{d x}$ (product rule).
If you differentiate with respect to $y$ instead, the factor $\frac{d x}{d y}$ appears.

✍️ Worked examples

Example 1 Find $\frac{d y}{d x}$ if $x^{2} + y^{2} = 25$ , then find the tangent and normal at $(3, 4)$ .
Solution:
Differentiate 2x+2ydydx=0.
So $\frac{d y}{d x} = - \frac{x}{y}$ .
At $(3, 4)$ the slope is $- \frac{3}{4}$ .
Tangent y−4=−34(x−3), that is 3x+4y=25.
Normal slope is the negative reciprocal, $\frac{4}{3}$ , so the normal is $y - 4 = \frac{4}{3} (x - 3)$ .
Example 2 Find $\frac{d y}{d x}$ for $x^{3} + y^{3} = 6 x y$ and evaluate it at $(3, 3)$ .
Solution:
Differentiate both sides 3x2+3y2dydx=6y+6xdydx.
Group (3y2−6x)dydx=6y−3x2.
So $\frac{d y}{d x} = \frac{2 y - x^{2}}{y^{2} - 2 x}$ .
At (3,3) 6−99−6=−1.
Example 3 Find $\frac{d y}{d x}$ if $x \sin y + y \cos x = 2$ .
Solution:
Differentiate sin⁡y+xcos⁡ydydx+cos⁡xdydx−ysin⁡x=0.
Group (xcos⁡y+cos⁡x)dydx=ysin⁡x−sin⁡y.
So $\frac{d y}{d x} = \frac{y \sin x - \sin y}{x \cos y + \cos x}$ .
Example 4 Find $\frac{d y}{d x}$ if $\tan 2 x + \cot y = x y$ .
Solution:
Differentiate 2sec⁡22x−csc⁡2ydydx=y+xdydx.
Group −(x+csc⁡2y)dydx=y−2sec⁡22x.
So $\frac{d y}{d x} = \frac{2 \sec^{2} 2 x - y}{x + \csc^{2} y}$ .
Example 5 Find $\frac{d y}{d x}$ if $x e^{y} = y + 1$ .
Solution:
Differentiate ey+xeydydx=dydx.
Group (1−xey)dydx=ey.
So $\frac{d y}{d x} = \frac{e^{y}}{1 - x e^{y}}$ .

Important remarks

The final formula usually contains both $x$ and $y$ .
To find the slope at a given $x$ you must first find the matching $y$ from the original equation.
A point must lie on the curve before you substitute it; check it in the equation.
A horizontal tangent needs the numerator of $\frac{d y}{d x}$ equal to zero; a vertical tangent needs the denominator equal to zero.

Implicit Differentiation, Parametric Differentiation and Higher Derivatives · 3 / 9

1-1 (c) Parametric differentiation

Parametric curves

If $x = f (t)$ and $y = g (t)$ with the same domain, the pair of equations describes one curve, and $t$ is the parameter.
Each value of $t$ gives one point $(x, y)$ of the curve.

Comparison aspect	Main idea	Keywords
•	If $f$ and $g$ are differentiable in $t$ and $\frac{d x}{d t} \neq 0$ , then $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$	differentiabledxdtthendydx
This is just the chain rule	dydt=dydx⋅dxdt	dydtdydxdxdt
Horizontal tangent	dydt=0 and dxdt≠0	dydtdxdt
Vertical tangent	dxdt=0 and dydt≠0	dxdtdydt

✍️ Worked examples

Example 6 For $x = t^{2} + 1$ and $y = t^{3} - 3 t$ find $\frac{d y}{d x}$ when $t = 2$ .
Solution:
$\frac{d x}{d t} = 2 t$ and $\frac{d y}{d t} = 3 t^{2} - 3$ .
$\frac{d y}{d x} = \frac{3 t^{2} - 3}{2 t}$ .
At t=2 12−34=94.
Example 7 For $x = 3 \cos 2 θ$ and $y = 4 \sin 3 θ$ find $\frac{d y}{d x}$ at $θ = \frac{π}{4}$ .
Solution:
$\frac{d x}{d θ} = - 6 \sin 2 θ$ and $\frac{d y}{d θ} = 12 \cos 3 θ$ .
$\frac{d y}{d x} = \frac{12 \cos 3 θ}{- 6 \sin 2 θ} = \frac{- 2 \cos 3 θ}{\sin 2 θ}$ .
At θ=π4 cos⁡3π4=−22 and sin⁡π2=1.
So $\frac{d y}{d x} = - 2 (- \frac{\sqrt{2}}{2}) = \sqrt{2}$ .
Example 8 Find the points where the curve $x = t^{3} - 3 t$ , $y = t^{2} - 4 t$ has a horizontal or vertical tangent.
Solution:
$\frac{d x}{d t} = 3 t^{2} - 3 = 3 (t - 1) (t + 1)$ and $\frac{d y}{d t} = 2 t - 4$ .
Horizontal tangent 2t−4=0 gives t=2, where dxdt=9≠0. The point is (2,−4).
Vertical tangent dxdt=0 gives t=±1, where dydt≠0 (it is −2 and −6).
The points are $(- 2, - 3)$ for $t = 1$ and $(2, 5)$ for $t = - 1$ .

Implicit Differentiation, Parametric Differentiation and Higher Derivatives · 4 / 9

1-1 (d) Derivative of one function with respect to another

The idea

To differentiate $y = f (x)$ with respect to $z = g (x)$ , treat $x$ as the parameter of both.
$\frac{d y}{d z} = \frac{d y / d x}{d z / d x}$

✍️ Worked examples

Example 9 Find the derivative of $4 x^{3} - 9 x^{2} + 5$ with respect to $3 x^{2} + 7$ .
Solution:
Let $y = 4 x^{3} - 9 x^{2} + 5$ and $z = 3 x^{2} + 7$ .
$\frac{d y}{d x} = 12 x^{2} - 18 x$ and $\frac{d z}{d x} = 6 x$ .
$\frac{d y}{d z} = \frac{12 x^{2} - 18 x}{6 x} = 2 x - 3$ .
Example 10 Find the derivative of $x - \sin x$ with respect to $1 - \cos x$ at $x = \frac{π}{3}$ .
Solution:
$\frac{d y}{d x} = 1 - \cos x$ and $\frac{d z}{d x} = \sin x$ .
$\frac{d y}{d z} = \frac{1 - \cos x}{\sin x}$ .
At x=π3 1−1232=13=33.

Implicit Differentiation, Parametric Differentiation and Higher Derivatives · 5 / 9

1-2 (a) Higher-order derivatives

Definition and notation

If $y = f (x)$ is differentiable, then $y^{'} = \frac{d y}{d x} = f^{'} (x)$ is a new function.
If $y^{'}$ is differentiable, its derivative is the second derivative: $y^{''} = \frac{d^{2} y}{d x^{2}} = f^{''} (x)$ .
Repeating gives the third derivative $y^{'''} = \frac{d^{3} y}{d x^{3}}$ and, in general, the $n$ th derivative $y^{(n)} = \frac{d^{n} y}{d x^{n}} = f^{(n)} (x)$ , with $n$ a positive integer.
Derivatives from the second onwards are called higher derivatives.

$\frac{d^{2} y}{d x^{2}}$ means the derivative of $\frac{d y}{d x}$ .
${(\frac{d y}{d x})}^{2}$ means the square of the first derivative.
They are different quantities; never mix them.

✍️ Worked examples

Example 11 Find the third derivative of $y = x^{4} - 2 x^{2} + 5$ .
Solution:
$y^{'} = 4 x^{3} - 4 x$ .
$y^{''} = 12 x^{2} - 4$ .
$y^{'''} = 24 x$ .
Example 12 Find the second derivative of $y = \frac{x + 1}{x - 1}$ .
Solution:
$y^{'} = \frac{(x - 1) - (x + 1)}{(x - 1)^{2}} = \frac{- 2}{(x - 1)^{2}} = - 2 (x - 1)^{- 2}$ .
$y^{''} = 4 (x - 1)^{- 3} = \frac{4}{(x - 1)^{3}}$ .
Example 13 Find the second derivative of $y = \sin (3 x - 2)$ and of $y = \sqrt{3 x - 2}$ .
Solution:
For the sine y′=3cos⁡(3x−2) and y′′=−9sin⁡(3x−2).
For the root y=(3x−2)1/2, so y′=32(3x−2)−1/2.
Then $y^{''} = \frac{3}{2} \cdot (- \frac{1}{2}) \cdot 3 (3 x - 2)^{- 3 / 2} = - \frac{9}{4} (3 x - 2)^{- 3 / 2}$ .

$y = x^{n}$ gives $y^{(n)} = n!$ , and every later derivative is $0$ .
$y = e^{k x}$ gives $y^{(n)} = k^{n} e^{k x}$ .
$y = \sin a x$ gives $y^{(n)} = a^{n} \sin (a x + \frac{n π}{2})$ ; for $y = \cos a x$ use $\cos (a x + \frac{n π}{2})$ .
Differentiating $\sin 2 x$ four times returns $16 \sin 2 x$ , so the pattern repeats every four steps.

Differentiate the given relation as many times as the required relation demands
then eliminate constants.

✍️ Worked examples

Example 14 If $y = \tan x$ prove that $y^{''} = 2 y (1 + y^{2})$ .
Solution:
$y^{'} = \sec^{2} x = 1 + \tan^{2} x = 1 + y^{2}$ .
$y^{''} = 2 y y^{'} = 2 y (1 + y^{2})$ .
Example 15 If $y = 3 \cos (2 x + 1)$ prove that $y^{''} + 4 y = 0$ .
Solution:
$y^{'} = - 6 \sin (2 x + 1)$ .
$y^{''} = - 12 \cos (2 x + 1) = - 4 (3 \cos (2 x + 1)) = - 4 y$ .
Hence $y^{''} + 4 y = 0$ .

Implicit Differentiation, Parametric Differentiation and Higher Derivatives · 6 / 9

1-2 (b) Second derivative of implicit relations

Method

Find $y^{'}$ first, simplify it, then differentiate again.
Replace any $y^{'}$ that appears by its formula, and use the original equation to simplify.

✍️ Worked examples

Example 16 If $x^{2} + y^{2} = 4$ prove that $y^{3} y^{''} + 4 = 0$ .
Solution:
$2 x + 2 y y^{'} = 0$ , so $y^{'} = - \frac{x}{y}$ .
$y^{''} = - \frac{y - x y^{'}}{y^{2}} = - \frac{y + \frac{x^{2}}{y}}{y^{2}} = - \frac{y^{2} + x^{2}}{y^{3}}$ .
Since $x^{2} + y^{2} = 4$ , $y^{''} = - \frac{4}{y^{3}}$ , so $y^{3} y^{''} + 4 = 0$ .
Example 17 If $y^{2} + 2 x y = 8$ find a relation containing $y^{''}$ , $y^{'}$ and $y$ .
Solution:
Differentiate 2yy′+2y+2xy′=0, so (x+y)y′+y=0.
Differentiate again (1+y′)y′+(x+y)y′′+y′=0.
Therefore $(x + y) y^{''} + (y^{'})^{2} + 2 y^{'} = 0$ .

Implicit Differentiation, Parametric Differentiation and Higher Derivatives · 7 / 9

1-2 (c) Second derivative of parametric curves

The rule

If x=f(t) and y=g(t) then d2ydx2=ddt(dydx)dxdt

First find $\frac{d y}{d x}$ as a function of $t$ , differentiate it with respect to $t$ , then divide by $\frac{d x}{d t}$ .
A frequent mistake is to write $\frac{d^{2} y / d t^{2}}{d^{2} x / d t^{2}}$ ; this is wrong.

✍️ Worked examples

Example 18 If $x = 2 n^{3} - 5$ and $y = 6 n^{2} + 1$ find $\frac{d^{2} y}{d x^{2}}$ at $n = 1$ .
Solution:
$\frac{d x}{d n} = 6 n^{2}$ and $\frac{d y}{d n} = 12 n$ .
$\frac{d y}{d x} = \frac{12 n}{6 n^{2}} = \frac{2}{n}$ .
$\frac{d}{d n} (\frac{2}{n}) = - \frac{2}{n^{2}}$ .
$\frac{d^{2} y}{d x^{2}} = \frac{- 2 / n^{2}}{6 n^{2}} = - \frac{1}{3 n^{4}}$ .
At $n = 1$ the value is $- \frac{1}{3}$ .
Example 19 For $x = \sec t$ , $y = \tan t$ show that $\frac{d^{2} y}{d x^{2}} = - \frac{1}{y^{3}}$ .
Solution:
$\frac{d x}{d t} = \sec t \tan t$ and $\frac{d y}{d t} = \sec^{2} t$ .
$\frac{d y}{d x} = \frac{\sec^{2} t}{\sec t \tan t} = \frac{\sec t}{\tan t} = \csc t$ .
$\frac{d}{d t} (\csc t) = - \csc t \cot t$ .
$\csc t \cot t = \frac{\cos t}{\sin^{2} t}$ and $\sec t \tan t = \frac{\sin t}{\cos^{2} t}$ .
$\frac{d^{2} y}{d x^{2}} = - \frac{\cos t}{\sin^{2} t} \cdot \frac{\cos^{2} t}{\sin t} = - \cot^{3} t$ .
Since $y = \tan t$ , $\cot^{3} t = \frac{1}{y^{3}}$ , so $\frac{d^{2} y}{d x^{2}} = - \frac{1}{y^{3}}$ .

Implicit Differentiation, Parametric Differentiation and Higher Derivatives · 8 / 9

1-2 (d) Meaning of the derivatives

Graph connection

$f^{'} (x)$ is the slope of the tangent; $f^{''} (x)$ tells how that slope changes.
Where $f^{''} (x) > 0$ the slope is increasing, so the graph bends upwards like a cup; where $f^{''} (x) < 0$ the slope is decreasing and the graph bends downwards like a cap.
For a polynomial, each differentiation lowers the degree by one, so $f^{'}$ and $f^{''}$ have simpler graphs than $f$ .
Plotting $f$ , $f^{'}$ and $f^{''}$ with graphing software shows that $f^{'}$ crosses the $x$ -axis at the turning points of $f$ , and $f^{''}$ crosses it at the points where the bending changes.

Point of comparison	Implicit differentiation	Parametric differentiation
Data given	One equation in $x$ and $y$	$x = f (t)$ , $y = g (t)$
First derivative	Differentiate both sides, solve for $y^{'}$	$\frac{d y / d t}{d x / d t}$
Answer contains	Usually $x$ and $y$	Only the parameter $t$
Second derivative	Differentiate $y^{'}$ again and substitute $y^{'}$	$\frac{d}{d t} (\frac{d y}{d x}) / \frac{d x}{d t}$

Implicit Differentiation, Parametric Differentiation and Higher Derivatives · 9 / 9

Curriculum

Mathematics

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

Implicit Differentiation, Parametric Differentiation and Higher Derivatives

🌳 Concept tree

├── 1-1 (a) Explicit and implicit relations

│ ├── Explicit functions

│ └── Implicit relations

├── 1-1 (b) Implicit differentiation

│ ├── The method

│ ├── The chain-rule step

│ └── Important remarks

├── 1-1 (c) Parametric differentiation

│ ├── Parametric curves

│ └── The rule

│ └── The idea

├── 1-2 (a) Higher-order derivatives

│ ├── Definition and notation

│ ├── A common confusion

│ ├── Patterns for the nth derivative

│ └── Proving relations between derivatives

│ └── Method

│ └── The rule

└── 1-2 (d) Meaning of the derivatives

├── Graph connection

└── Summary table

Explicit functions

The method

The chain-rule step

✍️ Worked examples

Important remarks

Parametric curves

✍️ Worked examples

The idea

✍️ Worked examples

Definition and notation

✍️ Worked examples

✍️ Worked examples

Method

✍️ Worked examples

The rule

If x=f(t) and y=g(t) then d2ydx2=ddt(dydx)dxdt

✍️ Worked examples

Graph connection

Well done! You finished this lesson