Maths with Lemon

Chain and the other Rules

"Love can reach the same level of talent, and even genius, as the discovery of differential calculus. "

Lev Vygotsky

Chain rule

What you have to know:

The derivatives of all known functions
(If not, check this lesson)
Recognize all the basic functions
You must be able to manipulate exponential expressions.
(If not, check this lesson)
You must be able to manipulate logarithmic expressions.
(If not, check this lesson)

Key Points

1. Watch the video:

2. Check out this presentation.
3. Watch this video.

Solved examples

Watch the videos

Chain rule for simple functions

1. Watch the video:

2. Watch this video.

Product Rule

What you have to know:

The derivatives of all known functions
(If not, check this lesson)
Recognize all the basic functions
You must be able to manipulate exponential expressions.
(If not, check this lesson)
You must be able to manipulate logarithmic expressions.
(If not, check this lesson)

Key Points

1. Watch the video:

2. Check out this presentation.
3. Watch this video.

Quotient Rule

What you have to know:

The derivatives of all known functions
(If not, check this lesson)
Recognize all the basic functions
You must be able to manipulate exponential expressions.
(If not, check this lesson)
You must be able to manipulate logarithmic expressions.
(If not, check this lesson)

Key Points

1. Watch the video:

2. Check out this presentation.
3. Watch this video.

IB questions on differentiation

What you have to know:

The derivatives of all known functions
(If not, check this lesson)
Recognize all the basic functions
You must be able to manipulate exponential expressions.
(If not, check this lesson)
You must be able to manipulate logarithmic expressions.
(If not, check this lesson)

Key Points

1. Watch the video:

2. Watch this video on derivatives and tangents.

Extra

Material and references:

Hodder Book SL(ISBN: 9781510462359) :
20A, 20B, 20C

Key Points

You should know how to use the chain rule to differentiate composite functions. If \( y = f(u) \), where \( u = g(x) \), then \[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}. \]

You should know how to differentiate products and quotients. If \( y = u(x)v(x) \), then \[ \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}. \] If \( y = \frac{u(x)}{v(x)} \), then \[ \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}. \]

IB Past Paper Problems

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