Exercise 6.06
The Chain Rule: Polynomials with brackets
In AS maths, we learnt learnt Simple Polynomial Differentiation...
...and the rule we had to adhere to was: get rid of any brackets before differentiating
Often it is too cumbersome to get rid of the brackets and sometimes it is plain impossible.
The Chain Rule let us differentiate expressions with brackets
- as long as all of the \(x\) s are inside the brackets
Web Lesson #23 showed you how to apply the Chain Rule:
There are two methods - mostly we can differentiate by sight - which was the method used in the movie
But, when we are stuck, we can revert back to FIRST PRINCIPLES:
The Chain Rule using FIRST PRINCIPLES
- Use the 'Chain Rule' whenever you have a set of brackets and ALL of the \(x\) s are inside the brackets
- If it isn't already written as \(y = \color{#9499b1}{...}\) then place a \(y=\) in-front of the expression (or an \(f\left( x \right) =\))
- Replace the expression in the brackets with \(U\) (Remember to state exactly what \(U\) stands for)
- So you now have TWO equations, one for \(y\) (in terms of \(U\) ), and the other for \(U\) (in terms of \(x\) )
- Differentiate each of these, to find \(\frac{dy}{dU}\) and \(\frac{dU}{dx}\)
- Multiply together the answers for \(\frac{dy}{dU}\) and \(\frac{dU}{dx}\) . Call this \(\frac{dy}{dx}\)
- Finally, substitute back to get rid of \(U\)
- Finally, check - numpties like you get a lot of crap wrong! YOU NEED TO CHECK-MATE! 🙄 sorry!
BE WARNED: If you rely too much on FIRST PRINCIPLES, you'll struggle later - when we get to integration!
e.g. If \(y=\left( 2x^2-7 \right) ^3\), find \(\frac{dy}{dx}\):
[Replace expression inside brackets]: y = U³ and U = 2x² - 7
[Differentiate both of these]: dy = 3U² and dU = 4x
dU dx
[Multiply together to get dy/dx]: dy = 12xU²
dx
[Substitute back]: dy = 12x(2x²-7)²
dx
e.g. Differentiate \(y=\frac{1}{3x^2-4x}\)
[Write as y = ...]: y = 1
(3x² - 4x)
[Re-write using indices]: y = (3x² - 4x)-1
[Replace expression inside brackets]: y = U-1 and U = 3x² - 4x
[Differentiate both of these]: dy = -U-2 and dU = 6x - 4
dU dx
[Multiply together to get dy/dx]: dy = -(6x-4)U-2
dx
[Substitute back]: dy = -(6x-4)(3x² - 4x)-2
dx
[Re-write in symbol form]: dy = - (6x-4)
dx (3x²-4x)²
[Simplify]: dy = 4-6x
dx (3x²-4x)²
e.g. Differentiate \(y=4x^2+\left( 2x^2-7 \right) ^3\)
Remember that we can differentiate term by term
We know (it's easy to differentiate 4x²) the differential of the 1st term is: 8x:
y = 4x² + (2x²-7)³
└┬┘ └─┬─┘ └────┬───┘
│ │ │
▼ ▼ ▼
dy = 8x + ......
dx
And, we know (from above) the differential of the second term is: 12x(2x²-7)²
y = 4x² + (2x²-7)³
└┬┘ └─┬─┘ └────┬───┘
│ │ │
▼ ▼ ▼
dy = 8x + 12x(2x²-7)²
dx
The animation in WL#23 shows how to apply the Chain Rule by sight
In these hints, I have used the slower first principles method
You need to be able to use BOTH methods
Question 1, part (a): STEP 1: VERIFY THAT THE CHAIN RULE APPLIES
╒════════════════════╕
Are all the xs in y = (x + 2)5
one set of brackets? └──┬──┘
╘════════════════════╛ └── ✔
STEP 2: DIFFERENTIATE:
You should be able to differentiate this By Sight (as shown in the the animation). You get this: \(\frac{dy}{dx}=5 \color{#EC4EEC}{\left( ... \right)} \left( ...+... \right) ^4\)
The bit in PINK is the differential of \(x+2\) - which is just \(\color{#EC4EEC}{1}\), so really, it doesn't even need to be there!
The method I'm showing here in the hints is the FIRST PRINCIPLES METHOD - which is really a fallback - and not needed here!
y = (x + 2)5 ╔═══════════════════════════╗
└──┬──┘ ▼ ╒═══════════╩══════════╕
┌─┴───────────┴────────┐ THIS IS 1st PRINCIPLES
▼ ┌───┴───┐ YOU SHOULDN'T USE THIS
Substituting the x+2: y = U5 and U = x+2 ╘══════════════════════╛
Differentiating these: dy = 5U4 and dU = 1
dU └─┬─┘ dx └┬┘
└───────────┬───────────┘
┌───┴───┐
Multiplying these together: dy = 1 Χ 5U4 = 5U4
dx └─┐
┌──┴─┐
Substituting back, we get: dy = 5(....)4
dx
STEP 3: USE THE DIFFERENTIATION BUTTON ON THE CALC TO CHECK!
Checking
The Casio FX991EX (classwiz) can check your ANSWER, so you're sure you got it right (or wrong!)
1) Put x = (say) 1.2 into your answer for \(\frac{dy}{dx}\) (this gives the gradient of the curve at x=1.2)
⬢ First store 1.2 -> x on your calculator:
┌────────── you don't press ALPHA
│ when STOring 1.2 as X
▼
On the calculator, TYPE: 1.2 STO x
⬢ Then put x=1.2 into your equation for \(\frac{dy}{dx}\) to get the gradient at x=1.2
TYPE: 5 ( x + 2 ) x□5 =
2) Use the \(\frac{d}{dx}\) function on your calculator with the original equation (for y) to find the gradient at x=1.2
TYPE: SHIFT d/dx ( x + 2 ) x□5 ⇨ │ₓ₌ 1.2 =
▲
Press the right ───────────┘ │
navigation-arrow to put x=1.2 │x=▢ ◄──── into the box
You can check every differential that you do...
...remember; if you ain't checkin', it means you is chicken 
Question 1, part (b): STEP 1: VERIFY THAT THE CHAIN RULE APPLIES
╒════════════════════╕
Are all the xs in y = x(x - 1)3
one set of brackets? ▲
╘════════════════════╛ └─────── ✘ What's this x doing out here? ✘
⛔⛔⛔ STOP! WE CAN'T USE THE CHAIN RULE ⛔⛔⛔
Just write N/A as the answer!
THE END
Question 1, part (c): STEP 1: VERIFY THAT THE CHAIN RULE APPLIES
╒════════════════════╕
Are all the xs in y = (3x - 1)7
one set of brackets? └───┬──┘
╘════════════════════╛ └── ✔
STEP 2: DIFFERENTIATE:
You should be able to differentiate this using the DIRECT METHOD (as shown in the the animation)...
You get this: \(\frac{dy}{dx}=7 \color{#EC4EEC}{\left( ... \right)} \left( ...+... \right) ^6\)
The bit in PINK is the differential of \(3x-1\) - which is just \(\color{#EC4EEC}{3}\)
The method I'm showing below is the FIRST PRINCIPLES - which is really a fallback - and not needed here!
y = (3x - 1)7 ╔═══════════════════════════╗
└───┬──┘ ▼ ╒═══════════╩══════════╕
┌──┴──────────┴────────┐ THIS IS 1st PRINCIPLES
▼ ┌───┴───┐ YOU SHOULDN'T USE THIS
Substituting the 3x-1: y = U7 and U = 3x-1 ╘══════════════════════╛
Differentiating these: dy = 7U6 and dU = 3
dU └─┬─┘ dx └┬┘
└───────────┬───────────┘
┌───┴───┐
Multiplying these together: dy = 3 Χ 7U6 =
U6
dx └─┐
┌──┴─┐
Substituting back: dy =
(3x-1)6
dx
STEP 3: USE THE DIFFERENTIATION BUTTON ON THE CALC TO CHECK!
Checking
The Casio FX991EX (classwiz) can check your ANSWER, so you're sure you got it right (or wrong!)
1) Put x = (say) 1.2 into your answer for \(\frac{dy}{dx}\) (this gives the gradient of the curve at x=1.2)
⬢ First store 1.2 -> x on your calculator:
┌────────── you don't press ALPHA
│ when STOring 1.2 as X
▼
On the calculator, TYPE: 1.2 STO x
⬢ Then put x=1.2 into your equation for \(\frac{dy}{dx}\) to get the gradient at x=1.2
TYPE: ⍰ ( 3 x - 1 ) x□6 =
2) Use the \(\frac{d}{dx}\) function on your calculator with the original equation (for y) to find the gradient at x=1.2
TYPE: SHIFT d/dx ( 3 x - 1 ) x□7 ⇨ │ₓ₌ 1.2 =
▲
Press the right ───────────┘ │
navigation-arrow to put x=1.2 │x=▢ ◄──── into the box
...remember; if you ain't checkin', it means you is chicken 
Question 1, part (d): STEP 1: VERIFY THAT THE CHAIN RULE APPLIES
╒════════════════════╕
Are all the xs in y = (4 - 3x)3
one set of brackets? └───┬──┘
╘════════════════════╛ └── ✔
STEP 2: DIFFERENTIATE:
You should be able to differentiate this using the DIRECT METHOD (as shown in the the animation)...
You get this: \(\frac{dy}{dx}=3 \color{#EC4EEC}{\left( ... \right)} \left( ...-... \right) ^{...}\)
The bit in PINK is the differential of \(4-3x\)
The method I'm showing here in the hints is the FIRST PRINCIPLES - which is really a fallback - and not needed here!
y = (4 - 3x)3 ╔═══════════════════════════╗
└───┬──┘ ▼ ╒═══════════╩══════════╕
┌──┴──────────┴────────┐ THIS IS 1st PRINCIPLES
▼ ┌───┴───┐ YOU SHOULDN'T USE THIS
Substituting the 3x-1: y = U3 and U = 4-3x ╘══════════════════════╛
Differentiating these: dy = 3U² and dU = -3
dU └─┬─┘ dx └┬─┘
└───────────┬───────────┘
┌───┴────┐
Multiplying these together: dy = ... Χ 3U
= -9U
dx
Substituting back, we get: dy = ........
dx
STEP 3: USE THE DIFFERENTIATION BUTTON ON THE CALC TO CHECK!
Question 1, part (g): There is NO NEED TO USEFIRST PRINCIPLES (which I'm showing you below). You should be able to differentiate this by sight (the DIRECT METHOD shown in the the animation)...
STEP 1: VERIFY THAT THE CHAIN RULE APPLIES
╒════════════════════╕
Are all the xs in y = (x² - x + 1)3
one set of brackets? └─────┬────┘
╘════════════════════╛ └── ✔
STEP 2: DIFFERENTIATE:
You should be able to differentiate this using the DIRECT METHOD (as shown in the the animation)...
You get this: \(\frac{dy}{dx}=3 \color{#EC4EEC}{\left( ...-... \right)} \left( ...-...+... \right) ^2\)
The bit in PINK is the differential of \(x^2-x+1\) - which is just \(\color{#EC4EEC}{2x-1}\)
The method I'm showing here in the hints is the FIRST PRINCIPLES - which is really a fallback - and not needed here!
y = (x² - x + 1)3 ╔═══════════════════════════╗
└───┬──┘ ▼ ╒═══════════╩══════════╕
┌──┴──────────┴────────┐ THIS IS 1st PRINCIPLES
▼ ┌───┴───┐ YOU SHOULDN'T USE THIS
Substituting the 3x-1: y = U3 and U = ... ╘══════════════════════╛
Differentiating these: dy = 3U² and dU = ...
dU └─┬─┘ dx └┬─┘
└───────────┬───────────┘
┌───┴────┐
Multiplying these together: dy = ... Χ 3U2 = 3(...)U2
dx
Substituting back, we get: dy = ........
dx
Don't forget to CHECK!
Question 1, part (i): STEP 1: DOES CHAIN RULE APPLY? It applies if there are brackets - BUT WHERE R THEY?
Of course, they are around the denominator of the fraction
╒════════════════════╕
Are all the xs in y = 1
one set of brackets? (x - 2)
╘════════════════════╛ └─────┬────┘
└── ✔
STEP 2: DIFFERENTIATE:
You should be able to differentiate this by sight (i.e. the DIRECT METHOD shown in the the animation)...
Using INDICES we get: \(y=\left( x -2 \right) ^{-1}\). Using the DIRECT METHOD, you should get: \(\frac{dy}{dx}=-1\color{#EC4EEC}{\left( ... \right)} \left( x-2 \right) ^{...}\)
The bit in PINK is the differential of the brackets, which is just \(\color{#EC4EEC}{1}\). Re-writing it in SYMBOL notation: \(\frac{dy}{dx}=\frac{...}{\left( x-2 \right) ^{2}}\)
The method I'm showing below is the FIRST PRINCIPLES - which is really a fallback - and not needed here!
y = 1
(x - 2) ╔═══════════════════════════╗
└───┬──┘ ▼ ╒═══════════╩══════════╕
┌──┴──────────┴────────┐ THIS IS 1st PRINCIPLES
▼ ┌───┴───┐ YOU SHOULDN'T USE THIS
Substituting the 3x-1: y = 1 ◄───┐ and U = ... ╘══════════════════════╛
U ◄───┤
┌──┴─┐
│ U⁻Ή │
└────┘
Differentiating these: dy = -U-2 and dU = ...
dU └─┬─┘ dx └┬─┘
└───────────┬───────────┘
┌───┴───────┐
Multiplying these together: dy = ... Χ -1/U² = -1/U²
dx
Substituting back, we get: dy = ........
dx
Don't forget to CHECK!
Question 1, part (j): STEP 1: DOES CHAIN RULE APPLY? It applies if there are brackets - BUT WHERE R THEY?
Of course, they are around the denominator of the fraction
┌───────┐
╒════════════════════╕ ▼ │
Are all the xs in y = x └──── ✘ What's this x doing out here? ✘
one set of brackets? (x - 2)
╘════════════════════╛
Question 1, part (k): Firstly, re-write it as: \(y=3\left( 1-2x \right) ^{-1}\)
Now, you should get the answer using the DIRECT METHOD (as shown in the lesson and in Web Lesson 23)...
...in case you got it wrong when you tried to do it using the DIRECT METHOD, these hints show you how to get the answer using FIRST PRINCIPLES
Replacing the '1-2x': y = 3 ◄───┐ and U = 1 - 2x
U ◄───┤
┌──┴───┐
│ 3U⁻Ή │
└──────┘
Differentiating these: dy = -U-2 and dU = ...
dU └─┬─┘ dx └┬─┘
└───────────┬───────────┘
┌───┴───────┐
Multiplying these together: dy = ... Χ -1/U² =
dx U²
Substituting back: dy =
dx (....)²
Question 1, part (m): Using the DIRECT METHOD, you should get something like this: \(\frac{dy}{dx}=4 \color{#EC4EEC}{\left( ...+\frac{...}{...} \right)} \left( x-\frac{1}{x} \right) ^3\)
The bit in PINK is the differential of the brackets!
Question 1, part (n): Firstly, re-write it using INDICES as: \(y=\left( x^2 +3 \right) ^{-1}\)
Using the DIRECT METHOD, you should get something like this: \(\frac{dy}{dx}=-1\color{#EC4EEC}{\left( ... \right)} \left( x^2+3 \right) ^{...}\)
The bit in PINK is the differential of the brackets!
The re-write it back in SYMBOL notation: \(\frac{dy}{dx}=\frac{...}{\left( x^2+3 \right) ^{...}}\)
Question 1, part (o): Firstly, re-write it using INDICES
Then, using the DIRECT METHOD, you should get something like this: \(\frac{dy}{dx}=-1 \color{#EC4EEC}{\left( ...+... \right)} \left(x^2 + x - 5 \right) ^{...}\)
Question 1, part (q): Firstly, re-write it using INDICES as: \(y=4\left( x -2 \right) ^{-2}\)
Using the DIRECT METHOD, you should get something like this: \(\frac{dy}{dx}=-2\color{#EC4EEC}{\left( ... \right)} \left( x -3 \right) ^{...}\)
The bit in PINK is the differential of the brackets!
The re-write it back in SYMBOL notation: \(\frac{dy}{dx}=\frac{...}{\left( x-2 \right) ^{...}}\)
Question 1, part (r): Using the DIRECT METHOD, you should get something like this: \(\frac{dy}{dx}=-3 \color{#EC4EEC}{\left(-...-... \right)} \left(... - ... - ... \right) ^{-4}\)
You can simplify this, but factorising the \(-\) out of the PINK brackets...
Question 1, part (s): Firstly, re-write it using INDICES as: \(y=\left( 2x-2x^{...} \right) ^3\)
Using the DIRECT METHOD, you should get something like this: \(\frac{dy}{dx}=3 \color{#EC4EEC}{\left( ...+... \right)} \left( ...-... \right) ^2\)
The re-write it back in SYMBOL notation...
Question 2, part (a): We apply the Chain Rule, when all \(x\)s are INSIDE-A-SET-OF-BRACKETS...
But here, all of the \(x\)s are inside-a-set-of-brackets, WHICH IS INSIDE ANOTHER SET OF BRACKETS!!!
These are question where we NEED to revert to the FIRST PRINCIPLES METHOD:
y = (1 + √½x-1)8
IDENTIFY THE OUTERMOST ───────────┐───────────────┐
SET OF BRACKETS
y = ⎧1 + (½x - 1)Ή⁄₂ ⎫8
⎩ ⎭
└─────────┬──────┘
┌────────┴─────────────┐
▼ ┌───┴───────────┐
Substituting: y = U⁸ and U = (½x - 1)Ή⁄₂
Differentiating these: dy = 8U⁷ and dU =
(½x - 1)⁻Ή⁄₂
dU └─┬─┘ dx └──────┬─────┘
└───────────┬─────────────────┘
┌───────────┴──────┐
Multiplying these together: dy =
(½x - 1)⁻Ή⁄₂ Χ 8U⁷ =
(½x - 1)⁻Ή⁄₂ U⁷
dx └──────┐
┌─────┴────┐
Substituting back, we get: dy =
(½x - 1)⁻Ή⁄₂ (1 + √½x-1)⁷
dx
STEP 3: CHECKING with \(x=1.2\) WON'T WORK - NEVER MIND, TRY \(x=2.4\) INSTEAD!!!
Question 2, part (b): Re-write (using indices) as:
\(y\,\,=\,\,\left( x+\left( 1-2x \right) ^4 \right) ^{-1}\)
Now search for the outermost brackets:
\(y\,\,=\,\,\color{#62be53}{\left( \color{#003D6D}{x+\left( 1-2x \right) ^4} \right) ^{\color{#003D6D}{-1}}}\)
and replace everything inside them with \(U\):
\(y=U^{-1}\) and \(U=x+\left( 1-2x \right) ^4\)
Differentiate each:
\(\frac{dy}{dU}=-U^{-2}\) and \(\frac{dU}{dx}=1-...\left( 1-2x \right) ^3\)
Multiply these and replace back the \(U\)...
Question 2, part (c): Re-write: \(y=\sqrt{1-\sqrt{1-4x}}\) using indices: \(y=\left( 1-\left( 1-4x \right) ^{1/2} \right) ^{1/2}\)
Now identify the outermost set of brackets: \(y=\color{#62be53}{\left( \color{#003D6D}{1-\left( 1-4x \right) ^{1/2}} \right)} ^{1/2}\) and replace the algebra inside those brackets with \(U\)
It's easy to differentiate: \(y=U^{1/2}\), but you need to take a bit of care differentiating: \(U\,\,=\,\,1-\left( 1-4x \right) ^{1/2}\)
Finally: CHECKING with \(x=1.2\) WON'T WORK - NEVER MIND, TRY \(x=0.12\) INSTEAD!!!
Question 3: We know that, on a displacement-time graph, the gradient gives the velocity...
differentiate differentiate
╔══════════►═►═►═►═►═══════════╦═══════════►═►═►═►═►═══════════╗
╒═══════╩═════╕ ╒══════╩═══════╕ ╒══════╩═══════╕
│displacement:│ │ velocity: │ │ acceleration:│
│ x │ │ v or dx │ │a or dv or d²x│
│ │ │ dt │ │ dt dt²│
╘═══════╦═════╛ ╘══════╦═══════╛ ╘══════╦═══════╛
╚══════════◄═◄═◄═◄═◄═══════════╩═══════════◄═◄═◄═◄═◄═══════════╝
integrate integrate
Question 4: If we have an equation for velocity, and we differentiate it, we get an equation for accelration:
differentiate differentiate
╔══════════►═►═►═►═►═══════════╦═══════════►═►═►═►═►═══════════╗
╒═══════╩═════╕ ╒══════╩═══════╕ ╒══════╩═══════╕
│displacement:│ │ velocity: │ │ acceleration:│
│ x │ │ v or dx │ │a or dv or d²x│
│ │ │ dt │ │ dt dt²│
╘═══════╦═════╛ ╘══════╦═══════╛ ╘══════╦═══════╛
╚══════════◄═◄═◄═◄═◄═══════════╩═══════════◄═◄═◄═◄═◄═══════════╝
integrate integrate
So, we need to differentiate: \(v=\left( 3t^2-t-1 \right) ^2\)
By now, you should be able to differentiate this with little NO problems...
So, \(\frac{dv}{dt}=2\left(
t-
\right) \left( 3t^2\;\;\;\;\;-\;\;t\;\;\;-\;\;1 \right) \)
dv = 2 (
t -
) (3t² - t - 1)
dt
When t=-1: dv = 2[
(-1) -
] [3(-1)² - (-1) - 1]
dt
dv = ...
dt
Question 5: Remember, when differentiating, we deal with ONE TERM AT A TIME:
Firstly, lets re-write this as: \(A=\color{#62be53}{\frac{1}{2}t^2}-\color{#2b83c3}{\frac{1}{5}\left( 1-t \right) ^3}\)
Now, when we differentiate the \(\color{#62be53}{\frac{1}{2}t^2}\), we get just: \(\color{#62be53}{t}\)
When we differentiate the \(\color{#2b83c3}{\frac{1}{5}\left( 1-t \right) ^3}\), we get: \(\color{#2b83c3}{
\left( 1-t \right) ^2}\)
If you're struggling to differentiate:\(\color{#2b83c3}{\frac{1}{5}\left( 1-t \right) ^3}\) using the DIRECT METHOD, you could always revert to FIRST PRINCIPLES, like a total baby 👶🍼
\(y_{\tiny{1}}=\color{#2b83c3}{\frac{1}{5} \color{#00ADB1}{U}^3}\) where \(\color{#00ADB1}{U=1-t}\) \(\Rightarrow \) \(\color{#2b83c3}{\frac{dy_{\tiny{1}}}{dU}=
\color{#00ADB1}{U}^2}\) and \(\color{#00ADB1}{\frac{dU}{dx}=-1}\) \(\Rightarrow \) \(\frac{dy}{dx}=..........\)
Simplify: dA = t +
(1-t)²
dt
This expands to: = t +
(1-t)(1-t)
= t +
(1 2t + t²)
= t +
⁶⁄₅t +
t²
Which simplifies to: =
- ⅕ t +
t²
Which factorises to: =
(
t² t +
)
Now, we want dA/dt to equal 1: 1 =
(
t² t +
)
Which solves to give: 5 =
t² t +
=> t =
and t =
Question 7: Differentiating one term at a time, we get: \(\frac{dy}{dx}=\color{#62be53}{2}+\color{#2b83c3}{\frac{
}{\left( x+1 \right) ^2}}\)
We want to find \(\frac{dy}{dx}\) (i.e. the gradient) when \(x = 3\). Putting in \(x = 3\), we get:
\(\frac{dy}{dx}\,\,=\,\,2 +\,\,\frac{...}{\left( \color{#006566}{\left( -3 \right)} +1 \right) ^2}\)
Which gives: \(\frac{dy}{dx}\,=\,...\)
Question 8: In this question \(U\) would equal \({x}^2 - x - 3\)
Question 9: Differentiating (using the chain rule), you get: \(\frac{dy}{dx}\,=\,\left( ......... \right) ^{-3/2}\)
We want to know where \(\frac{dy}{dx} = \frac{1}{8}\)
So: \(\left( ......... \right) ^{-3/2}\,\,=\,\,\color{#bd398c}{\frac{1}{8}}\)
Solving this, we get \(x=...\)
Now, we also need the y-coordinate
So, putting (\(x=...\)) into the equation for \(y\), we get: \(y=...\)
Question 10: We'll need to use the Chain Rule in either case; but it might be easier to just expand that denominator to get: \(y=\frac{1}{4x-x^2}\)
Differentiating: \(\frac{dy}{dx}=\frac{\left( ...-... \right)}{\left( 4x-x^2 \right) ^2}\)
The turning points are when \(\frac{dy}{dx}\,=\,0\)
Multiplying through by: \(\left( 4x-x^2 \right) ^2\), it becomes really simple to solve...
To determine the nature of the roots, we have two choices:
1) Use the gradient before and after method ◄──┐
└─we put in a number a bit less
2) Use the second derivative ◄────┐ (0.1 less) than the root
┌───────────────────┐ │ then a bit more (0.1 more)
│ we'll struggle to └─┬─┘ than the root
│ differentiate dy/dx └──┐
│ because the Chain Rule └────┐
│ doesn't apply. All the xs └───┐
│ AREN'T inside 1=set of brackets │
└─────────────────────────────────┘
Question 13: Differentiating the first term is easy:
y = 2x - 1
d│ 3 + μx
i│ └───┬──┘
f│ ┌───┴───────────────────────┐
f│ │ y = 1/U and U = 3+μx │
r│ │ dy = -1 and dU = μ │
n│ │ dU U² dx │
t│ └───┬───────────────────────┘
▼ ┌───┴──┐
dy = 2 - │ │
dx └──────┘
If you are unsure how I differentiated 3+μx, I thought of it as 3+10x for a second. I realised that that would differentiate to give 10, so 3+μx would differentiate to give μ
We are told there is a turning point at x = 1; so: when x = 1, dy/dx = 0
Putting these in, we get: 0 = 2 - μ
[3+μ(1)]²
This simplifies:
= - μ
(3+μ)²
Multiplying through by (3+a)², we get:
(3+μ)² = -μ
Multiplying out the (3+a)(3+a) first:
(
+
μ + μ²) = -μ
Now, multiplying out the '2' we get:
+
μ +
μ² = -μ
Which gives:
μ² +
μ +
= 0
Which factorises: (
+
)(
+
) = 0
Which gives: => μ =
and μ =
Question 15: We have THREE unknowns and we know THREE pieces of information about this curve...
1) We know that, when \(x=0\), \(L=\frac{1}{4}\)
2) We know that, when \(x=1\), \(\frac{dy}{dx}=0\)
3) We know that, when \(x=1\), \(L=-2\)
(1) - we can plonk this straight into the equation, and that will give up \(a\) right away...
(2) - we need to differentiate first and then stuff these values in; but on it's own, this equation won't tell or \(b\) or \(c\)
(3) - so we need to insert these values into the original equations and think simultaneous
Question 17: Sneakily, they haven't given the \(x\)-value where they want the tangent/normal; they've given the \(y\)-value!
So, let: \(\frac{2}{1-2x}\) equal \(1\) and solve...
Okay - so we know the coordinates of a point on our tangent; but we need to know it's gradient - so differentiate: \(\frac{dy}{dx}=\frac{...}{\left( 1-2x \right) ^2}\)...
...and insert the value of \(x\) we calculated...
(NOTE: You can check you got the correct gradient by asking your calculator to work out \(\frac{d}{dx}\left[ \frac{2}{1-2x} \right] _{|x=-\frac{1}{2}}\))
Use: \(\frac{d}{dx}\left[ \frac{2}{1-2x} \right] _{|x=-\frac{1}{2}}\) to get the tangent ...
Then, change the sign and recip the gradient and use \(\frac{d}{dx}\left[ \frac{2}{1-2x} \right] _{|x=-\frac{1}{2}}\) again to find the normal!
Question 20: This is an implicit equation (\(y\) ain't the subject), but we could easily re-write it as: \(y=\sqrt{8+x^2}\) (did I forget to use \(\pm \)?)
Because \(P\) lies on the curve, it's y-coordinate is given by \(\pm \)
So, we can say that \(P\) has coordinates: \(P=\left( x, \sqrt{8+x^2} \right) \)
And we know A has coordinates: \(A=\left( 2, 0 \right) \)
So we can find an equation for the length: \(\left| \overrightarrow{AP} \right|\)
P = ( x, √8 + x² )
A = ( 2, 0 )
----------------------------
│AP│² = (x-2)² + (√8 + x²)²
│AP│ = √2x²
Now we want to MINIMISE the value of \(\left| \overrightarrow{AP} \right|\), which is OPTIMISATION: We diifferentiate, make it = 0 and solve...
|