Maths C3 - Differentiation... Help??

OP

My first question is about the Chain rule...

Do I have to learn this version??...
Chain Rule.png

Or can I simply learn the other version which is...

\frac{dy}{dx}= \frac{dy}{du} \times \frac{du}{dx}

??

I'm not going to lie, I don't understand anything from the first set of formulas. As far as I'm aware, not even ExamSolutions covers it! :frown:

(edited 7 years ago)

Here we go again!

OP

Original post by RDKGames

Here we go again!

Just when you thought you heard the last of me :P

Original post by Philip-flop

My first question is about the Chain rule...

Do I have to learn this version??...
Chain Rule.png

Or can I simply learn the other version which is...

\frac{dy}{dx}= \frac{dy}{du} \times \frac{du}{dx}

??

I'm not going to lie, I don't understand anything from the first set of formulas. As far as I'm aware, not even ExamSolutions covers it! :frown:

You can remember the last one because the ones above it have been derived by applying the one you know to certain forms of functions without loss of generality. I'll show you how your

\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}

fits into the first two definitions. Though it tells you to learn them, I never did. You just get used to them as you practice them, and you can always derive them as I will show you below.

1. As an example, pick

y=(x^2+1)^2

then we let

u=x^2+1 \Rightarrow \frac{du}{dx}=2x

and we also now have

y=u^2 \Rightarrow \frac{dy}{du}=2u

So, coming from the rule you know, we say

\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx} = 2u \cdot 2x = 4x(x^2+1)

from simplifying and reverting back to

x

from

u

Now let's generalise this:

Say I have a function

y=[f(x)]^n

. This would link to our example above because

f(x)=x^2+1

and

n=2

.

Now what did we do? We set whatever is inside the bracket to some new variable, so we let

u=f(x)\Rightarrow \frac{du}{dx}=f'(x)

as required. And we are now left with

y=u^n

which we know how to differentiate with respect to u so we get

\frac{dy}{du}=n\cdot u^{n-1}

Following, again, from the equation you know we get:

\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} = nu^{n-1} \cdot f'(x)=n[f(x)]^{n-1}f'(x)

as shown in the book.

2. As an example, pick

f(x)=2x^2+1

and

g(x)=x^4-x^3+1

then we can let

y=fg(x)

which you should recognise as a composite function. It means the exact same thing as

y=f[g(x)]

just different notation.

So we have

y=fg(x)=2(x^4-x^3+1)^2+1

and we want to differentiate this.

Now we let

u=x^4-x^3+1 \Rightarrow \frac{du}{dx}=4x^3-3x^2

We also have

y=2u^2+1 \Rightarrow \frac{dy}{du}=4u

So,

\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} = 4u \cdot (4x^3-3x^2) = 16x^3(x^4-x^3+1)-12x^2(x^4-x^3+1)

Now let's generalise it.

We have

y=f[g(x)]

and we let

u=g(x) \Rightarrow \frac{du}{dx}=g'(x)

And

y=f(u) \Rightarrow \frac{dy}{du}=f'(u)

So;

\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} = f'(u) \cdot g'(x) = f'[g(x)] \cdot g'(x)

as shown in the book.

:smile:

(edited 7 years ago)

Reply 5

OP

Original post by RDKGames

You can remember the last one because the ones above it have been derived by applying the one you know to certain forms of functions without loss of generality. I'll show you how your

\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}

fits into the first two definitions. Though it tells you to learn them, I never did. You just get used to them as you practice them, and you can always derive them as I will show you below.
...
...
:smile:

Wow, thanks for that brilliant explanation!! I may have to read it a few more times to let everything fully sink in though! But this definitely makes more sense to me now! Thank you @RDKGames :smile:

Reply 6

OP

Ok so here is my noob question of the day :colondollar:

....

I'm currently watching a video on Exam Solutions about 'The Product Rule'...
The Product Rule.png

... But I've hit a brick wall. How do I differentiate the term

Unparseable latex formula:

e^3^x

... as in

\frac{dv}{dx}

part?

Original post by Philip-flop

Ok so here is my noob question of the day :colondollar:

....

I'm currently watching a video on Exam Solutions about 'The Product Rule'...
The Product Rule.png

... But I've hit a brick wall. How do I differentiate the term

Unparseable latex formula:

e^3^x

... as in

\frac{dv}{dx}

part?

v=e^{3x} \Rightarrow \frac{dy}{dx}=3e^{3x}

Use the chain rule with

g=3x

if you can't differentiate

e^{3x}

all in one go

Reply 8

Original post by Philip-flop

My first question is about the Chain rule...

Do I have to learn this version??...
Chain Rule.png

Or can I simply learn the other version which is...

\frac{dy}{dx}= \frac{dy}{du} \times \frac{du}{dx}

??

I'm not going to lie, I don't understand anything from the first set of formulas. As far as I'm aware, not even ExamSolutions covers it! :frown:

If you had something like

(3x^2+3)^4

, you can differentiate that using the chain rule with the longwinded approach of making a substitution.

Or with practice you can do it the quick way in your head using the formula for the derivative of anything of the form

f(x)^n

. And this extends to any composition of functions

fg(x)

.

Now you've had an explanation, can you differentiate

(3x^2+3)^4

without using a substitution?

(edited 7 years ago)

Reply 9

OP

Original post by RDKGames

v=e^{3x} \Rightarrow \frac{dy}{dx}=3e^{3x}

Use the chain rule with

g=3x

if you can't differentiate

e^{3x}

all in one go

I still don't think I understand :frown:

I know that to differentiate you have to do

\frac{dy}{dx} = na^{n-1}

but I don't know how to use it on the

e^{3x}

taking your advice... I'm not sure I know how to apply the chain rule to it either :/

Original post by Philip-flop

I still don't think I understand :frown:

I know that to differentiate you have to do

\frac{dy}{dx} = na^{n-1}

but I don't know how to use it on the

e^{3x}

taking your advice... I'm not sure I know how to apply the chain rule to it either :/

y=e^{3x}

Let

u=3x \Rightarrow \frac{du}{dx}=3

also

y=e^u \Rightarrow \frac{dy}{du}=e^u

So

\frac{dy}{dx}=...

Reply 11

OP

Original post by notnek

If you had something like

(3x^2+3)^4

, you can differentiate that using the chain rule with the longwinded approach of making a substitution.

Or with practice you can do it the quick way in your head using the formula for the derivative of anything of the form

f(x)^n

. And this extends to any composition of functions

fg(x)

.

Now you've had an explanation, can you differentiate

(3x^2+3)^4

without using a substitution?

Ok so...

If...

y=(3x^2+3)^4

then...

f(x) = 3x^2+3

To differentiate using the chain rule...

[f(x)]^n = n[f(x)]^{n-1}f'(x)

this gives ....

4[f(x)]^3f'(x)

4(3x^2+3)^3(6x)

Am I right?

Reply 12

Original post by Philip-flop

Ok so...

If...

y=(3x^2+3)^4

then...

f(x) = 3x^2+3

To differentiate using the chain rule...

[f(x)]^n = n[f(x)]^{n-1}f'(x)

this gives ....

4[f(x)]^3f'(x)

4(3x^2+3)^3(6x)

Am I right?

Correct. Basically the chain rule is used when you have a function of a function. In reality that just means that there are brackets somewhere. So all of these can be differentiated using the chain rule:

(3x^2+2)^3

e^{(3x)}

\sin(x^2)

Chain rule in words : Make a substitution for the stuff in the brackets then differentitate what you have now and then multiply by the derivative of the stuff in the brackets.

So

(3x^2+2)^3

:

Start by differentiating

t^3

which is

3t^2

then multiply by the derivative of

3x^2+2

.

So you're left with

3(3x^2+2)^2 \times 6x

Next use the same method for

e^{(3x)}

:

First differentiate

e^t

which is

e^t

(this is one of the standard derivatives that you'll need to learn).

Then multiply this by the derivative of

3x

.

So you get

e^{3x} \times 3

.

There are standard results that are useful to learn e.g. the derivative of

e^{ax}

is

ae^{ax}

. But you can derive all these quickly using the chain rule.

(edited 7 years ago)

Reply 13

OP

Original post by RDKGames

y=e^{3x}

Let

u=3x \Rightarrow \frac{du}{dx}=3

also

y=e^u \Rightarrow \frac{dy}{du}=e^u

So

\frac{dy}{dx}=...

Oh right I see!

I now understand that...

y = ae^x

when differentiating gives...

\frac{dy}{dx} = ae^x

... where a is a constant.

I think you've cleared this up now. Thanks for the help :smile:

Reply 14

OP

Original post by notnek

Correct. Basically the chain rule is used when you have a function of a function. In reality that just means that there are brackets somewhere. So all of these can be differentiated using the chain rule:

(3x^2+2)^3

e^{(3x)}

\sin(x^2)

Chain rule in words : Make a substitution for the stuff in the brackets then differentitate what you have now and then multiply by the derivative of the stuff in the brackets.

So

(3x^2+2)^3

:

Start by differentiating

t^3

which is

3t^2

then multiply by the derivative of

3x^2+2

.

So you're left with

3(3x^2+2)^2 \times 6x

Next use the same method for

e^{(3x)}

:

First differentiate

e^t

which is

e^t

(this is one of the standard derivatives that you'll need to learn).

Then multiply this by the derivative of

3x

.

So you get

e^{3x} \times 3

.

There are standard results that are useful to learn e.g. the derivative of

e^{ax}

is

ae^{ax}

. But you can derive all these quickly using the chain rule.

Thank you @notnek!! I'm still having trouble with the derivatives of exponential, natural logs, and trigonometric types. I know how to get there but I just don't fully understand how they are derived (if that makes sense?). For now I will just have to remember them and keep practising until everything clicks..

Differentiation Exponential, Log, and Trig types.png

Differentiation Exponential, Log, and Trig types.png

Reply 15

Original post by Philip-flop

Thank you @notnek!! I'm still having trouble with the derivatives of exponential, natural logs, and trigonometric types. I know how to get there but I just don't fully understand how they are derived (if that makes sense?). For now I will just have to remember them and keep practising until everything clicks..

Differentiation Exponential, Log, and Trig types.png

If you want help with these types then post some example questions.

Reply 16

OP

Original post by notnek

If you want help with these types then post some example questions.

I just don't understand how the examples in the red boxes above differentiate to what they are, which is making it more difficult to remember them. But I will attempt some more questions on these early tomo morning and see whether I have any more questions :smile:

Reply 17

Original post by Philip-flop

I just don't understand how the examples in the red boxes above differentiate to what they are, which is making it more difficult to remember them. But I will attempt some more questions on these early tomo morning and see whether I have any more questions :smile:

Oh I see. All of these are standard results that you need to learn for C3 but don't need to know where they come from.

There is no rule e.g. the chain rule to derive these results. They can be derived from first principles of calculus. You should google them if you want more explanation but you only need to quote them for C3 without proof.

You do however need to use the chain rule with these standard results. E.g. you need to be able to find the derivative of something like

\sin \left( x^2\right)

using the fact that the derivative of

\sin x

is

\cos x

.

EDIT : The derivative of

\tan x

can be derived from the derivatives of

\sin x

and

\cos x

. You may be asked to show this in a C3 exam.

(edited 7 years ago)

Reply 18

OP

Original post by notnek

Oh I see. All of these are standard results that you need to learn for C3 but don't need to know where they come from.

There is no rule e.g. the chain rule to derive these results. They can be derived from first principles of calculus. You should google them if you want more explanation but you only need to quote them for C3 without proof.

You do however need to use the chain rule with these standard results. E.g. you need to be able to find the derivative of something like

\sin \left( x^2\right)

using the fact that the derivative of

\sin x

is

\cos x

.

EDIT : The derivative of

\tan x

can be derived from the derivatives of

\sin x

and

\cos x

. You may be asked to show this in a C3 exam.

That's good then!!

Thanks again

Oh really? I may need to make note of how tan x is derived from the derivatives of sin x and cos x then. Once I figure out how how it is derived :tongue:

Reply 19