Differentiating 20x(2x-1)^6 with product and chain rules

[The Bob]

Quick question.

To differentiate $$y = 20x (2x - 1)^6$$ I need to use the product rule:

$$y = uv \ \Rightarrow \ \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}$$

So that u = 20x and v = (2x-1)⁶

However, do I differentiate v so that it equals 6(2x-1)⁵ or do I use:

$$y = [f(x)]^n \ \Rightarrow \ \frac{dy}{dx} = n[f(x)]^{n-1} f'(x)$$

to give 12(2x-1)⁵?

Cheers.

The Bob (2004 ©)

 

[arildno]

Why should it equal 6(2x-1)^5?
You are to differentiate v with respect to "x", not with respect to "2x-1".

 

[The Bob]

Why should it equal 6(2x-1)^5?
You are to differentiate v with respect to "x", not with respect to "2x-1".

Therefore it must be 12(2x-1)⁵?

The Bob (2004 ©)

 

[arildno]

Precisely, by the chain rule!

 

[The Bob]

Precisely, by the chain rule!

Cheers.

Sorry it seemed so trivial but I only started teaching myself tonight and the book I have sometimes misses some of the stages out e.g. I actually have v = (x-1)⁶ but this will come to 6(x-1)⁵ as the x-1 will become 1.

Was simply checking.

Appreciate the help.

Cheers again.

The Bob (2004 ©)

 

[dextercioby]

It would be much easier to put Maple/other software to expand that bynomial make the multiplications and the then differentiate each term of the resulting sum.

My guess...

Daniel.

 

[arildno]

It would be much easier to put Maple/other software to expand that bynomial make the multiplications and the then differentiate each term of the resulting sum.

My guess...

Daniel.

Are you trying to bait me?