Question

derivative of (2x + 1)^3

Answer 1

is the 2x considered an internal function, if so, I would multiply 3(2x + 1)2 by 2?

Answer 2

Example:

\(\large\color{black}{ y= (4x + 1)^4 }\)

the derivative would be:

\(\large\color{black}{ y'= 4(4x + 1)^{4-1} \times (4x+1)' }\)

\(\large\color{black}{ y'= 4(4x + 1)^{3} \times 4}\)

\(\large\color{black}{ y'= 16(4x + 1)^{3}}\)

Answer 3

@cleetus779 You are correct

Answer 4

Yes, that is correct. Good job!

Answer 5

Here is my result:\[
d(2x + 1)^3 = du^3 = 3u^2du = 3(2x+1)^2d(2x+1) = 6(2x+1)^2dx\\
\implies \frac{d(2x+1)^3}{dx} = 6(2x+1)^2
\]

Answer 6

to me it looks even a bit more confusing using a "u"... but everyone has a different taste.

Answer 7

many thanks to all, the problem actually asks for the derivative of tan^2((2x + 1)^3), can you help me continue?

Answer 8

Ohh, sure: Can you tell me what is the derivative of tan(x) ?

Answer 9

I get the 'u' stuff, it actuallyhelps to visualize it differently, so I appreciate that

Answer 10

what is the derivative of tan(x) ?

Answer 11

sec^2(x)

Answer 12

yes.

Answer 13

now product rule?

Answer 14

So what you are getting is.

\(\large\color{black}{ \tan[~(2x + 1)^3 ~] }\)
the derivative would be:

\(\large\color{black}{ \sec^2~[~(2x + 1)^3 ~] \times 3(2x + 1)^2 \times 2 }\)

Answer 15

see how there is there multiple chains going on?

Answer 16

processing

Answer 17

(you are using chain rule twice)

Answer 18

going to work out solution, will post what i get, could you do the same?

Answer 19

I already have the solution

Answer 20

anyway, work it out, and we will see if anything should be corrected....