No, because it isn’t sine multiplied by zero and cosine multiplied by zero — it is sine OF zero and cosine OF zero.

Sine and cosine are real-valued functions. They take in numbers as inputs and produce real number outputs.

When you input zero into sine the output is zero.

But when you input zero into cosine the output is one.

Therefore

.

Regards,

J.P.

My question is about p. 112 Q.2.

When I differentiate and all is well. But when shouldn’t the substitution everything equal to zero?

Because anything multiplied by zero is zero?

So if I substitute 0 shouldn’t this all be 0?

?

Regards.

]]>I am not altogether sure what you mean by “substitution after that”.

So, the chain rule says that

In words, the derivative of a composition is the derivative of the ‘outside’ – evaluated at the inside – times the derivative of the inside.

Here the outside is and the inside is . The derivative of the outside is , evaluated at the inside is , and the derivative of the inside is . Therefore we get

.

As it happens this differentiation is in the tables:

.

Altogether this gives, via the produce rule,

.

Regards,

J.P.

Just wondering if you could help with the differentiation of .

I understand to use the product rule to start, but then when I apply the chain rule to the I get confused with the substitution after that. Could you help.

Regards and Thanks.

]]>The confusion here, largely my fault perhaps, is that when you see in the calculus context that it is actually the natural log. So that, in this context:

.

This is the confusion.

So what you are differentiating is . It isn’t and so you have a chain rule where the ‘outside’ is and the ‘inside’ is .

The chain rule says that

,

that is “differentiate the outside — evaluate that at the inside — and the multiply by the derivative of the inside’.

The derivative of the outside, is in the tables and is . Evaluating this at the inside gives but we must multiply this by the derivative of the inside, the derivative of , which is .

All in all we have:

.

You probably need to move beyond this revision and tackle the new material — it might be the case that you don’t have to calculate more difficult derivatives such as these to do the new material.

Regards,

J.P.

Stuck on p. 102, Q 1 (f)

I have attached my attempt. I’ve been checking out khan academy and it said I needed the change of base formula but I got confused from there.

Thanks.

]]>