Your work in general is very good. I will focus on the questions you mentioned here.

P. 111, Q. 1 (d)

My advice would be to differentiate the two parts, , and , separately before using the product rule.

You do the chain rule slightly different to how I do the chain rule. I will do it your way and then my way, and then explain why I possibly prefer my way.

Give or take, your method is to identify and and then say that:

.

This is perfectly OK. My only concern is that I am not sure how you would get on if also needs a chain rule. Say if you had to differentiate:

.

Well anyway, using your method:

where . Using now your formula:

.

How do I do it? I use:

.

I do not necessarily label/name (outside function), and (inside function). Here is “the derivative of the outside function evaluated at the inside function”.

Here the outside function is and the inside function is so I get:

.

Both my method and your method are valid.

O.K., back to:

. Now we identify, for a product rule:

and we use the product rule (but we have already found the derivative of separately. So we get:

,

which looks different to the answer in the book but is equal to it.

I think you just went a little wrong by not taking your time.

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p.111, Q. 2 (a)

First of all your

is correct.

This gives the slope of the tangent to the point at ANY point. But you are interested in the point , i.e. . So you need to evaluate this at :

.

You can put this in the calculator, or note that and are an inverse pair so that:

.

At any rate .

Now you have

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p. 123, Q. 3

You have done all the hard work to get to

.

There is two approaches here.

1. Note that a fraction if and only if (and .

Proof: Suppose . Suppose . Multiply both sides by :

That is doing to the dog on the proof a little.

2. Or argue like this directly. Look at

What is making this difficult? The divides by . What is the inverse of dividing by ? Multiplying by . Do this to both sides to get:

(because ).

Now solve . I will let you take it from here.

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P.124, Q. 7

I can’t see where you have tried this so not sure where you are having difficulty.

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P. 124, Q. 8

You don’t have to calculate the second derivative. It won’t be examinable.

I can’t see how you would have a problem with the first derivatives.

_____________________________________________________

Please get back to me if you have any questions.

Regards,

J.P.

Struggling with a couple of exercises: 1(d) and 2(a) on p.111; Q. 3,7 and 8 on p123.

Thanks

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